Entrance Corner

Hint

(c) We know, speed of sound in a medium \(v=\sqrt{\frac{B}{\rho}}\) where \(\mathrm{B}=\) bulk modulus, \(\rho=\) density of the medium
Since, solids and liquids are much more difficult to compress than gases so they have much higher values of bulk modulus.
\(
\text { i.e., } B_{\text {solid }}>B_{\text {liquid }}>B_{\text {gas }}
\)
Generally solids and liquids have higher mass densities \((\rho)\) than gases. But corresponding increase in bulk modulus is much higher.
So, \(v_{\text {solid }}>v_{\text {liquid }}>v_{\text {gas }}\)

Hint

(d) Given, the equation of the transverse wave: \(y(x, t)=4.0 \sin \left[20 \times 10^{-3} x+600 t\right] \mathrm{mm}\)
\(x\) is in mm
\(t\) is in seconds

The general form of a transverse wave equation is \(y(x, t)=A \sin (k x \pm \omega t+\phi)\), where:
\({A}\) is the amplitude
\(k\) is the wave number
\(\omega\) is the angular frequency
\(\phi\) is the phase constant

The wave velocity \(v\) is given by \(v=\frac{\omega}{k}\).
If the sign in front of \(\omega t\) is positive, the wave travels in the negative \(x\)-direction. If the sign is negative, the wave travels in the positive x -direction.

How to solve? velocity using the formula \(v=\frac{\omega}{k}\).
Step 1: Identify \(\omega\) and \(k\) from the given equation
Comparing \(y(x, t)=4.0 \sin \left[20 \times 10^{-3} x+600 t\right]\) with the general form \(y(x, t)=A \sin (k x \pm \omega t+\phi)\), we get:
\(k=20 \times 10^{-3} \mathrm{~mm}^{-1}\)
\(\omega=600 \mathrm{~s}^{-1}\)

Step 2: Calculate the wave velocity \(v\)
Using the formula \(v=\frac{\omega}{k}\) :
\(v=\frac{600 \mathrm{~s}^{-1}}{20 \times 10^{-3} \mathrm{~mm}^{-1}}\)
\(v=\frac{600}{20 \times 10^{-3}} \frac{\mathrm{~mm}}{\mathrm{~s}}\)
\(v=30000 \frac{\mathrm{~mm}}{\mathrm{~s}}\)
\(v=30 \frac{\mathrm{~m}}{\mathrm{~s}}\)

Step 3: Determine the direction of the wave
Since the sign in front of \(\omega t\) is positive, the wave travels in the negative x-direction. Therefore, the velocity is \(-30 \mathrm{~m} / \mathrm{s}\).
The velocity of the wave is \(-30 \mathrm{~m} / \mathrm{s}\).

Hint

(b) Step 1: Frequency Formula for Organ Pipes
Closed Pipe (Odd Harmonics Only):
\(
f_n=\frac{n v}{4 L_c}, \quad n=1,3,5,7, \ldots
\)
Open Pipe (All Harmonics):
\(
f_n=\frac{m v}{2 L_o}, \quad n=1,2,3,4, \ldots
\)
Speed of Sound in a Gas:
\(
v=\sqrt{\frac{B}{\rho}}
\)
For closed pipe (9th harmonic, \(n=9\) ):
\(
f_9=\frac{9 v_1}{4 L_c}
\)
For open pipe (4th harmonic, \(n=4\) ):
\(
f_4=\frac{4 v_2}{2 L_o}=\frac{2 v_2}{L_o}
\)
Since the two frequencies are equal:
\(
\frac{9 v_1}{4 L_c}=\frac{2 v_2}{L_o}
\)

Step 2: Relating Speeds Using Density Ratio
Since both gases have the same bulk modulus, the speed of sound ratio is:
\(
\frac{v_1}{v_2}=\sqrt{\frac{\rho_2}{\rho_1}}=\sqrt{\frac{16}{1}}=4
\)
Thus, \(v_1=4 v_2\).
Substituting \(v_1=4 v_2\) in the equation:
\(
\begin{gathered}
\frac{9\left(4 v_2\right)}{4 L_c}=\frac{2 v_2}{L_o} \\
L_o=\frac{2 L_c}{9}
\end{gathered}
\)
Given \(L_c=10 \mathrm{~cm}\) :
\(
L_o=\frac{2 \times 10}{9}=\frac{20}{9} \mathrm{~cm}
\)

Hint

(c) The equation of the wave: \(y=2 \cos 2 \pi(330 t-x) m\).

The general form of a wave equation is \(y=A \cos (2 \pi f t-k x)\), where \(f\) is the frequency and \(k\) is the wave number.

How to solve?
Compare the given equation with the general form of a wave equation to find the frequency.

Step 1: Compare the given equation with the general form

The given equation is \(y=2 \cos 2 \pi(330 t-x)\).
The general form is \(y=A \cos (2 \pi f t-k x)\).
Comparing the two equations, we have \(2 \pi f=2 \pi \cdot 330\).

Step 2: Solve for the frequency \(f\)
Divide both sides of the equation by \(2 \pi\) :
\(\frac{2 \pi f}{2 \pi}=\frac{2 \pi \cdot 330}{2 \pi}\)
\(f=330 \mathrm{~Hz}\)

The frequency of the wave is 330 Hz.

Hint

(a) Given, Length of the open pipe: \(L_{\text {open }}=60 \mathrm{~cm}\)
The fundamental frequency of the closed pipe equals the first overtone frequency of the open pipe.
Fundamental frequency of a closed pipe: \(f_{\text {clased }}=\frac{v}{4 L_{\text {clased }}}\)
First overtone frequency of an open pipe: \(f_{\text {open }}=\frac{2 v}{2 L_{\text {open }}}=\frac{v}{L_{\text {open }}}\)

How to solve?
Equate the fundamental frequency of the closed pipe to the first overtone frequency of the open pipe and solve for the length of the closed pipe.

Equate frequencies
Set the fundamental frequency of the closed pipe equal to the first overtone frequency of the open pipe:
\(f_{\text {closed }}=f_{\text {open }}\)
\(\frac{v}{4 L_{\text {clased }}}=\frac{v}{L_{\text {open }}}\)
\(
L_{\text {closed }}=\frac{L_{\text {open }}}{4}
\)
\(
L_{\text {closed }}=\frac{60 \mathrm{~cm}}{4}=15 \mathrm{~cm}
\)
The length of the closed pipe is 15 cm.

Hint

(b) Given, The equation of motion: \(y=\sin (\omega t)+\cos (\omega t)\).

Trigonometric identity: \(\sin (a+b)=\sin a \cos b+\cos a \sin b\).
\(
\sin \left(\frac{\pi}{4}\right)=\cos \left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} .
\)

How to solve?
Rewrite the equation using the trigonometric identity for \(\sin (a+b)\) and then identify the amplitude.

Step 1: Rewrite the equation using the trigonometric identity
Multiply and divide the equation by \(\sqrt{2}\) :
\(y=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin (\omega t)+\frac{1}{\sqrt{2}} \cos (\omega t)\right)\)

Substitute \(\frac{1}{\sqrt{2}}\) with \(\sin \left(\frac{\pi}{4}\right)\) and \(\cos \left(\frac{\pi}{4}\right)\) :
\(y=\sqrt{2}\left(\sin (\omega t) \cos \left(\frac{\pi}{4}\right)+\cos (\omega t) \sin \left(\frac{\pi}{4}\right)\right)\)

Apply the trigonometric identity \(\sin (a+b)=\sin a \cos b+\cos a \sin b\) :
\(y=\sqrt{2} \sin \left(\omega t+\frac{\pi}{4}\right)\)

Step 2: Identify the amplitude

The amplitude of the motion is \(\sqrt{2}\).

Hint

\(
\text { (c) Speed of transverse wave }=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{70}{7 \times 10^{-3}}}=100 \mathrm{~m} / \mathrm{s}
\)

Hint

(d)

\(
\begin{aligned}
&\because y(x, t)=[0.05 \sin (8 x-4 t)] \mathrm{m}\\
&\begin{aligned}
\text { Speed of wave } & =\left|\frac{\text { Coefficient of } t}{\text { Coefficient of } x}\right| \\
& =\frac{4}{8}=0.5 \mathrm{~ms}^{-1}
\end{aligned}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& v_{\text {wave }}=\left|\frac{\text { coefficient of } t }{\text { coefficient of } x}\right| \\
& =\frac{400}{1}=400 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(a) The wave velocity is given by \(v=\frac{\omega}{k}\).
The maximum particle velocity is \(v_{p, \max }=A \omega\).
From the given equation, the amplitude \(A=2 \mathrm{~cm}\).
Therefore, \(v_{p, \max }=2 \omega\).
Equate the wave velocity and maximum particle velocity

Set \(v=v_{p, m a x}\)
\(\frac{\omega}{k}=2 \omega\)
\(
k=\frac{1}{2}
\)
Use the relationship \(k=\frac{2 \pi}{\lambda}\) :
\(\frac{1}{2}=\frac{2 \pi}{\lambda}\)
\(\lambda=4 \pi\)
The wavelength for which the wave velocity equals the maximum particle velocity is \(4 \pi \mathrm{~cm}\).

Hint

(d) We know, the equation of wave travelling in positive \(x\)-direction is 
\(
y=A \sin (k x-\omega t)
\)
where \(k=\frac{2 \pi}{\lambda}\)
and \(\omega=2 \pi f\)
Here given \(\lambda=4 \mathrm{~cm}\) and frequency \(({f})=100 \mathrm{~Hz}\)
\(
\therefore k=\frac{2 \pi}{4}=0.5 \pi \mathrm{~cm}^{-1}
\)
and \(\omega=2 \pi \times 100=200 \pi \mathrm{~s}^{-1}\)
\(\therefore\) Equation of travelling wave,
\(
y=A \sin (0.5 \pi x-200 \pi t)
\)

Hint

(b) Maximum particle velocity \(\omega A=(2 \pi n)(10)\)
Particle velocity = \(\left|\frac{\text { Coefficient of } t}{\text { Coefficient of } x}\right|\)
\(=n/(\frac{1}{\lambda})=n \lambda\)
\(
\text { given relationship } v_{p_{\max }}=4 v_w
\)
\(
\begin{aligned}
&(2 \pi n)(10)=(n \lambda)(4)\\
&\lambda=5 \pi
\end{aligned}
\)

Hint

(d) The speed of sound (\(v\)) is related to its frequency (\(f\)) and wavelength \((\lambda)\) by the formula: \({v}={f} \lambda\)

Beats are the result of the difference in frequencies between two sound waves.
Wavelength \(1(\lambda_1)=4.08 \mathrm{~m}\)
Wavelength \(2(\lambda_2)=4.16 \mathrm{~m}\)
Number of beats \((\mathrm{n})=40\) in 12 seconds, so the beat frequency (f_beat) \(=40 / 12 \mathrm{~Hz}\)
\(
\begin{aligned}
&\text { The beat frequency is the difference between the frequencies of the two waves: }\\
&\text { f_beat }=|f_1-f_2|
\end{aligned}
\)
\(
\begin{aligned}
& \frac{40}{12}=\frac{v}{4.08}-\frac{v}{4.16} \\
& \Rightarrow \frac{10}{3}=v\left(\frac{0.08}{4.08 \times 4.16}\right) \Rightarrow v=707.2 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(c) When a wave refracts into a denser medium, wavelength and speed decreases but frequency remains constant.

Explanation:
Refraction occurs when a wave enters a medium with a different density, causing it to change its speed and direction. In a denser medium, the wave slows down. Since the wave’s speed and wavelength are related ( \({v}={f} \lambda\) ), if the speed decreases, the wavelength must also decrease to maintain the same frequency. However, the frequency of the wave depends on the source and remains constant. This relationship is summarized by the equation: \({v}={f} \lambda\), where \(v\) is speed, \(f\) is frequency, and \(\)\lambda\(\) is wavelength.

Why other options are incorrect:
a: wavelength, speed and frequency decreases:
While the wavelength and speed decrease, the frequency does not. The frequency is determined by the source of the wave and stays constant.

b: wavelength increases, speed decreases and frequency remains constant:
When entering a denser medium, the wavelength decreases, not increases. As explained above, the wavelength and speed are inversely proportional, so if the speed decreases, the wavelength must decrease to maintain the constant frequency.

d: wavelength, speed and frequency increases:
This option is incorrect. In a denser medium, the wave’s wavelength and speed decrease, not increase. The frequency remains constant.

Hint

(a)

\(
\begin{aligned}
& y_1=5 \sin (2 \pi x-2 \pi v t) \\
& y_2=3 \sin (2 \pi x-2 \pi v t+3 \pi) \\
& \Rightarrow \text { Phase difference } \phi=3 \pi \\
& \Rightarrow A_{\text {net }}=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos (3 \pi)} \\
& \Rightarrow A_{\text {net }}=2 \mathrm{~cm}
\end{aligned}
\)

Hint

(c) Given, Equation of motion: \(x(t)=A \sin (\omega t)+B \cos (\omega t)\)
Angular frequency: \(\omega=\sqrt{\frac{K}{m}}\)
Displacement equation: \(x(t)=C \cos (\omega t-\phi)\)
Initial position: \({x}({0})\)
Initial velocity: \(v(0)\)
How to solve?
Find \(A\) and \(B\) using initial conditions.
Express \(C\) and \(\phi\) in terms of \(A\) and \(B\).
Substitute \(A\) and \(B\) in terms of \(x(0)\) and \(v(0)\) into the expressions for \(C\) and \(\phi\).

Step 1: Find \({A}\) and \({B}\) using initial conditions

At \(t=0, x(0)=A \sin (0)+B \cos (0)\).
Since \(\sin (0)=0\) and \(\cos (0)=1\), we have \(x(0)=B\).
The velocity \(v(t)\) is the derivative of \(x(t)\) with respect to \(t\) :
\(v(t)=\frac{d}{d t}(A \sin (\omega t)+B \cos (\omega t))=A \omega \cos (\omega t)-B \omega \sin (\omega t)\).
At \(t=0, v(0)=A \omega \cos (0)-B \omega \sin (0)\).
Since \(\cos (0)=1\) and \(\sin (0)=0\), we have \(v(0)=A \omega\).
Therefore, \(A=\frac{v(0)}{\omega}\) and \(B=x(0)\).

Step 2: Express \(C\) and \(\phi\) in terms of \(A\) and \(B\)
Expand \(x(t)=C \cos (\omega t-\phi)\) using the trigonometric identity:
\(x(t)=C(\cos (\omega t) \cos (\phi)+\sin (\omega t) \sin (\phi))\).
Comparing this with \(x(t)=A \sin (\omega t)+B \cos (\omega t)\), we get:
\(A=C \sin (\phi)\) and \(B=C \cos (\phi)\).
Squaring and adding these equations:
\(A^2+B^2=C^2 \sin ^2(\phi)+C^2 \cos ^2(\phi)=C^2\left(\sin ^2(\phi)+\cos ^2(\phi)\right)=C^2\).
Thus, \(C=\sqrt{A^2+B^2}\).
Dividing the equations:
\(\frac{A}{B}=\frac{C \sin (\phi)}{C \cos (\phi)}=\tan (\phi)\).
Thus, \(\phi=\arctan \left(\frac{A}{B}\right)\).

Step 3: Substitute \({A}\) and \({B}\) in terms of \(x(0)\) and \(v(0)\) into the expressions for \(C\) and \(\phi\)
Substituting \(A=\frac{v(0)}{\omega}\) and \(B=x(0)\) into \(C=\sqrt{A^2+B^2}\) :
\(C=\sqrt{\left(\frac{v(0)}{\omega}\right)^2+(x(0))^2}=\sqrt{\frac{v(0)^2}{\omega^2}+x(0)^2}\).
Substituting \(A=\frac{v(0)}{\omega}\) and \(B=x(0)\) into \(\phi=\arctan \left(\frac{A}{B}\right)\) :
\(\phi=\arctan \left(\frac{\frac{\nu(0)}{\omega}}{x(0)}\right)=\arctan \left(\frac{v(0)}{\omega x(0)}\right)\).

\(C\) and \(\phi\) are \(\sqrt{\frac{v(0)^2}{\omega^2}+x(0)^2}\) and \(\arctan \left(\frac{v(0)}{\omega x(0)}\right)\) respectively.

Alternate:

\(
\begin{aligned}
&C \cos \phi=x(0)\\
&t C \omega \sin \phi=v(0)\\
&\left[\frac{v(0)}{\omega}\right]^2+[x(0)]^2=C^2\\
&\tan \phi=\frac{v(0)}{x(0) \omega}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& Y=A \sin (k x-\omega t) \\
& A=\frac{6}{2}=3 \mathrm{~cm}=0.03 \mathrm{~m} \\
& \omega=2 \pi f=2 \pi \times 245 \\
& \omega=1.5 \times 10^3 \\
& k=\frac{\omega}{v}=\frac{1.5 \times 10^3}{300} \\
& k=5.1 \\
& y=0.03 \sin \left(5.1 x-\left(1.5 \times 10^3\right) t\right)
\end{aligned}
\)

Hint

(a) Frequency of known fork: \(f_1=340 \mathrm{~Hz}\)
Initial beat frequency: \(b_1=5\) beats/s
Beat frequency after filing: \(b_2=2\) beats/s

Beat frequency is the absolute difference between two frequencies: \(b=\left|f_1-f_2\right|\).
Filing a tuning fork increases its frequency.

How to solve?
Determine the initial possible frequencies of fork \(A\), then use the information about filing the fork to deduce the correct frequency.

Step 1: Calculate the possible frequencies of fork A before filing.
The beat frequency is given by \(b_1=\left|f_A-f_1\right|\).
So, \(5=\left|f_A-340\right|\).
This gives two possibilities: \(f_A=340+5=345 \mathrm{~Hz}\) or \(f_A=340-5=335 \mathrm{~Hz}\).

Step 2: Analyze the effect of filing.
Filing the fork increases its frequency.
If the initial frequency was 345 Hz , filing it would increase the frequency, moving it further away from 340 Hz , and increasing the beat frequency, which contradicts the problem statement.
If the initial frequency was 335 Hz , filing it would increase the frequency, moving it closer to 340 Hz , and decreasing the beat frequency, which agrees with the problem statement.

Step 3: Determine the correct frequency.
Since the beat frequency decreased after filing, the initial frequency of fork A must have been 335 Hz .

Hint

(d) What’s given in the problem
Diameter of the column tube: \({d}={6} \mathbf{~ c m}\)
Frequency of the tuning fork: \(f=504 \mathrm{~Hz}\)
Speed of sound: \(v=336 \frac{\mathrm{~m}}{\mathrm{~s}}\)

The formula for the wavelength of the sound wave at first resonance is \(\lambda=4(l+0.3 d)\), where \(l\) is the length of the air column and \(d\) is the diameter of the tube.
The relationship between speed, frequency, and wavelength is \(v=f \lambda\).

How to solve
Calculate the wavelength using the speed and frequency, then use the wavelength to find the length of the air column at first resonance, considering the end correction.

Step 1: Calculate the wavelength of the sound wave
Use the formula \(v=f \lambda\).
Rearrange the formula to solve for \(\lambda: \lambda=\frac{v}{f}\).
Substitute the given values: \(\lambda=\frac{336 \frac{\mathrm{~m}}{\mathrm{~s}}}{504 \mathrm{~Hz}}\).
Calculate the wavelength: \(\lambda=\frac{2}{3} \mathrm{~m}=66.67 \mathrm{~cm}\).

Step 2: Calculate the end correction
Use the formula for end correction: \({e}=\mathbf{0 . 3 d}\).
Substitute the given diameter: \(e=0.3 \times 6 \mathrm{~cm}\).
Calculate the end correction: \(c=1.8 \mathrm{~cm}\).

Step 3: Calculate the length of the air column at first resonance

Use the formula for the first resonance: \(\lambda=4(l+e)\).
Rearrange the formula to solve for \(l: l=\frac{\lambda}{4}-{e}\).
Substitute the calculated values: \(l=\frac{66.67 \mathrm{~cm}}{4}-1.8 \mathrm{~cm}\).
Calculate the length: \(l=16.67 \mathrm{~cm}-1.8 \mathrm{~cm}=14.87 \mathrm{~cm}\).

The reading of the water level in the column when the first resonance occurs is approximately 14.87 cm .

Alternate:

\(
\begin{aligned}
&\therefore l+1.8=\frac{\lambda}{4}\\
&\text { Also } \lambda=\frac{v}{f}=\frac{336}{504}\\
&\Rightarrow l+1.8=\frac{336}{4 \times 504}\\
&\Rightarrow l=14.86 \mathrm{~cm}
\end{aligned}
\)

Hint

(c) A travelling wave can be represented by the general form \(y(x, t)=A \sin (k x \pm \omega t+\phi)\), where \(A\) is the amplitude, \(k\) is the wave number, \(\omega\) is the angular frequency, \(t\) is time, \(x\) is position, and \(\phi\) is the phase constant.
Step 1: Analyze option a
\(y=A e^x \cos (\omega t-\theta)\) is not a travelling wave because of the \(e^x\) term, which indicates an exponential change in amplitude with position.

Step 2: Analyze option b
\(y=A e^{-x^2}(v t+\theta)\) is not a travelling wave because of the \(e^{-x^2}\) term, which indicates a Gaussian amplitude profile, and the term \((v t+\theta)\) is not in the form of \(k x \pm \omega t\).

Step 3: Analyze option c
\(y=A \sin (15 x-2 t)\) is a travelling wave because it matches the general form \(A \sin (k x-\omega t)\), where \(k=15\) and \(\omega=2\).

Step 4: Analyze option d
\(y=A \sin x \cos \omega t\) is not a travelling wave because it represents a standing wave, which can be seen by using the trigonometric identity
\(2 \sin a \cos b=\sin (a+b)+\sin (a-b)\) to rewrite it as \(y=\frac{A}{2}[\sin (x+\omega t)+\sin (x-\omega t)]\), a
superposition of two waves travelling in opposite directions.

The equation that represents a travelling wave is \(y=A \sin (15 x-2 t)\).

Hint

(b) Relationship between pressure difference, density, speed of sound, angular frequency and displacement: \(p=\rho v \omega s\)
\(
s=\frac{p}{\rho v \omega}
\)
\(
\begin{aligned}
&\begin{gathered}
\Delta \mathrm{P}=\mathrm{BK}_0 \\
B=\text { Bulk modulus }=\rho \mathrm{V}^2 \\
\Rightarrow \Delta \mathrm{P}=\frac{\rho \mathrm{V}^2 2 \pi \mathrm{f} \mathrm{~S}_0}{\mathrm{~V}} \quad [\because K=\frac{\omega}{V}=\frac{2 \pi f}{V}] \\
\Rightarrow \mathrm{S}_0=\frac{\Delta \mathrm{P}}{\rho \mathrm{~V} 2 \pi \mathrm{f}}
\end{gathered}\\
&\text { Take multiplicative constant ” } 2 \pi \text { ” as } 1\\
&\Rightarrow \mathrm{S}_0=\frac{10}{1 \times 300 \times 1000}=\frac{1}{30} \mathrm{~mm}=\frac{3}{100} \mathrm{~mm}
\end{aligned}
\)

Hint

(a) Height of water for first resonance: \({h}_1=17.0 \mathrm{~cm}\)
Height of water for second resonance: \(h_2=24.5 \mathrm{~cm}\)
Velocity of sound in air: \(v=330 \frac{\mathrm{~m}}{\mathrm{~s}}\)
The distance between two consecutive resonances in a resonance tube is half the wavelength: \(\Delta h=\frac{\lambda}{2}\).

The relationship between velocity, frequency, and wavelength is: \(v=f \lambda\).
How to solve?
Calculate the wavelength using the difference in height between the two resonances, then calculate the frequency using the wave equation.
Step 1: Calculate the difference in height between the two resonances.
\(
\begin{aligned}
& \Delta h=h_2-h_1 \\
& \Delta h=24.5 \mathrm{~cm}-17.0 \mathrm{~cm} \\
& \Delta h=7.5 \mathrm{~cm}
\end{aligned}
\)

Step 2: Calculate the wavelength.
\(
\begin{aligned}
& \Delta h=\frac{\lambda}{2} \\
& \lambda=2 \Delta h \\
& \lambda=2 \cdot 7.5 \mathrm{~cm} \\
& \lambda=15 \mathrm{~cm}
\end{aligned}
\)
Convert to meters: \(\lambda=0.15 \mathrm{~m}\)

Step 3: Calculate the frequency.
\(
\begin{aligned}
& v=f \lambda \\
& f=\frac{v}{\lambda} \\
& f=\frac{330 \frac{\mathrm{~m}}{\mathrm{~s}}}{0.15 \mathrm{~m}} \\
& f=2200 \mathrm{~Hz}
\end{aligned}
\)
The frequency of the tuning fork is 2200 Hz.

Note: In a resonance tube, the difference in height between two consecutive resonances corresponds to half the wavelength \((\lambda / 2)\).

Hint

(b) The distance between two consecutive crests is one wavelength \((\lambda)\).
The distance between a crest and a trough is half a wavelength \(\left(\frac{\lambda}{2}\right)\).
How to solve?
Set up equations based on the given information and solve for possible wavelengths.

Step 1: Set up the equation for the distance between two crests.
The distance between two crests is equal to \(n \lambda\), where \(n\) is an integer and \(\lambda\) is the wavelength.
\(
n \lambda=5
\)
Step 2: Set up the equation for the distance between a crest and a trough.
The distance between a crest and a trough is equal to \(\left(m+\frac{1}{2}\right) \lambda\), where \(m\) is an integer.
\(
\left(m+\frac{1}{2}\right) \lambda=1.5
\)
Step 3: Solve for \(\lambda\) in both equations.
From the first equation:
\(\lambda=\frac{5}{n}\)
From the second equation:
\(\lambda=\frac{1.5}{m+\frac{1}{2}}=\frac{3}{2 m+1}\)

Step 4: Equate the two expressions for \(\lambda\).
\(
\frac{5}{n}=\frac{3}{2 m+1}
\)
Step 5: Rearrange the equation.
\(
\begin{aligned}
& 5(2 m+1)=3 n \\
& 10 m+5=3 n
\end{aligned}
\)
Step 6: Find integer solutions for \(m\) and \(n\).
If \(m=1\), then \(10(1)+5=15=3 n\), so \(n=5\).
\(\lambda=\frac{5}{5}=1\)

If \(m=4\), then \(10(4)+5=45=3 n\), so \(n=15\).
\(\lambda=\frac{5}{15}=\frac{1}{3}\)

If \(m=7\), then \(10(7)+5=75=3 n\), so \(n=25\).
\(\lambda=\frac{5}{25}=\frac{1}{5}\)

Step 7
List the possible wavelengths.
The possible wavelengths are \(1 \mathrm{~m}, \frac{1}{3} \mathrm{~m}, \frac{1}{5} \mathrm{~m}\), and so on.

The possible wavelengths are \(1 \mathrm{~m}, \frac{1}{3} \mathrm{~m}, \frac{1}{5} \mathrm{~m}\), etc.

Hint

(a)

\(
\mathrm{T}_1=2 \mathrm{~g}
\)
\(
\mathrm{T}_2=8 \mathrm{~g}
\)
\(
\mathrm{V}=\sqrt{\frac{T}{\mu}}
\)
\(
\therefore \mathrm{V} \propto \sqrt{T}
\)
Also \(V=f \lambda\)
\(
\therefore \mathrm{v}_1=\mathrm{f}_1 \lambda_1
\)
and \(V_2=f_2 \boldsymbol{\lambda}_2\)
We know frequency of sources are same.
\(
\therefore \mathrm{f}_1=\mathrm{f}_2
\)
So \(\sqrt{\frac{T_1}{T_2}}=\frac{\lambda_1}{\lambda_2}\)
\(
\begin{aligned}
& \Rightarrow \sqrt{\frac{2 g}{8 g}}=\frac{6}{\lambda_2} \\
& \Rightarrow \lambda_2=12 \mathrm{~cm}
\end{aligned}
\)

Explanation for alternate solution: Length of the rope: \(L=12 \mathrm{~m}\)
Mass of the rope: \(m_{\text {rope }}=6 \mathrm{~kg}\)
Mass of the block: \(m_{\text {Hock }}=2 \mathrm{~kg}\)
Wavelength at the bottom: \(\lambda_{\text {batom }}=6 \mathrm{~cm}=0.06 \mathrm{~m}\)
The velocity of a wave on a string is given by \(v=\sqrt{\frac{T}{\mu}}\), where \(T\) is the tension and \(\mu\) is the linear mass density.
The linear mass density is given by \(\mu=\frac{m}{L}\).
The relationship between velocity, wavelength, and frequency is \(v=f \lambda\).
The frequency of the wave remains constant as it travels up the rope.

How to solve?
Find the ratio of the velocities at the top and bottom of the rope, which is equal to the ratio of the wavelengths, and then solve for the wavelength at the top.

Step 1: Calculate the tension at the bottom of the rope.
The tension at the bottom is due to the weight of the block:
\(T_{\text {bottom }}=m_{\text {blockg }}\)
\(T_{\text {bottom }}=2 \mathrm{~kg} \cdot 9.8 \frac{\mathrm{~m}}{\mathrm{~s}^2}=19.6 \mathrm{~N}\)

Step 2: Calculate the tension at the top of the rope.
The tension at the top is due to the weight of the block and the rope:
\(T_{\text {top }}=\left(m_{\text {block }}+m_{\text {rope }}\right) g\)
\(T_{\text {top }}=(2 \mathrm{~kg}+6 \mathrm{~kg}) \cdot 9.8 \frac{\mathrm{~m}}{\mathrm{~s}^2}=78.4 \mathrm{~N}\)

Step 3: Calculate the ratio of the velocities at the top and bottom.
Since \(v=\sqrt{\frac{T}{\mu}}\) and \(\mu\) is constant, the ratio of velocities is:
\(\frac{v_{\text {top }}}{v_{\text {bottom }}}=\sqrt{\frac{T_{\text {top }}}{T_{\text {bottom }}}}\)
\(\frac{v_{\text {top }}}{v_{\text {bottom }}}=\sqrt{\frac{78.4 \mathrm{~N}}{19.6 \mathrm{~N}}}=\sqrt{4}=2\)

Step 4: Calculate the wavelength at the top of the rope.
Since \(v=f \lambda\) and \(f\) is constant, the ratio of velocities is equal to the ratio of wavelengths:
\(\frac{v_{\text {top }}}{v_{\text {bottom }}}=\frac{\lambda_{\text {top }}}{\lambda_{\text {bottom }}}\)
\(\lambda_{\text {top }}=\lambda_{\text {bottom }} \cdot \frac{v_{\text {top }}}{v_{\text {bottom }}}\)
\(\lambda_{\text {top }}=0.06 \mathrm{~m} \cdot 2=0.12 \mathrm{~m}\)
\(\lambda_{\text {top }}=12 \mathrm{~cm}\)

The wavelength of the wavetrain when it reaches the top of the rope is 12 cm.

Hint

(a)

\(
\begin{aligned}
&\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}\\
&\text { For identical string } l \text { and } \mu \text { will be same }\\
&\begin{aligned}
& \mathrm{f} \propto \sqrt{T} \\
& \therefore \frac{450}{300}=\sqrt{\frac{T_x}{T_y}} \\
& \Rightarrow \frac{T_x}{T_y}=\frac{9}{4}=2.25
\end{aligned}
\end{aligned}
\)

Hint

(b) The difference in the frequencies of two successive harmonics is equal to fundamental frequency \(f=490-420=70 \mathrm{~Hz}\)
\(
\begin{aligned}
& f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \\
\Rightarrow & 70=\frac{1}{2 \times L} \sqrt{\frac{540}{6 \times 10^{-3}}} \\
\Rightarrow & L=\frac{1}{140} \times 300=2.142 \mathrm{~m}
\end{aligned}
\)

Hint

(a) Represent each wave as a phasor. Each wave has amplitude \(A\), where \(I_0 \propto A^2\). The phasors are:
\(
\begin{aligned}
& A_1=A \angle 0^{\circ}=A \\
& A_2=A \angle 45^{\circ}=A\left(\cos \left(45^{\circ}\right)+i \sin \left(45^{\circ}\right)\right)=A\left(\frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2}\right) \\
& A_3=A \angle-45^{\circ}=A\left(\cos \left(-45^{\circ}\right)+i \sin \left(-45^{\circ}\right)\right)=A\left(\frac{\sqrt{2}}{2}-i \frac{\sqrt{2}}{2}\right)
\end{aligned}
\)
Sum the phasors. The resultant phasor is \(A_R=A_1+A_2+A_3=A(1+\sqrt{2})\)
Calculate the resultant intensity. The intensity is proportional to the square of the amplitude. Therefore, the resultant intensity \(I_R\) is:
\(
I_R \propto\left|A_R\right|^2=A^2(1+\sqrt{2})^2=A^2(1+2 \sqrt{2}+2)=A^2(3+2 \sqrt{2}) \approx 5.8 A^2
\)
Since \(I_0 \propto A^2, I_R \approx 5.8 I_0\)

Hint

(d)

\(
\begin{aligned}
& v=\sqrt{\frac{T}{\mu}} \\
& \therefore \frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}} \\
& {v}_1={v}, {v}_2=\frac{v}{2} \\
& \Rightarrow \frac{v}{\frac{v}{2}}=\sqrt{\frac{2.06 \times 10^4}{T_2}} \\
& \Rightarrow \mathrm{~T}_2=\frac{2.06 \times 10^4}{4}=5.15 \times 10^3 \mathrm{~N}
\end{aligned}
\)

Hint

(a) Velocity of the wave, \({v}=\sqrt{\frac{T}{\mu}}\)
\(
\Rightarrow \mathrm{T}={v}^2 \mu
\)
We know, Youngs modulus,
\(
\mathrm{Y}=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}=\frac{\frac{T}{A}}{\frac{\Delta l}{l}}
\)
[As here \({F}={T}\) ]
\(
\begin{aligned}
& \Rightarrow Y \frac{\Delta l}{l}=\frac{T}{A}=\frac{v^2 \mu}{A} \\
& \Rightarrow \Delta l=\frac{v^2 \mu l}{A Y} \\
& =\frac{90 \times 90 \times \frac{60 \times 10^{-3}}{60 \times 10^{-2}} \times 60 \times 10^{-2}}{1 \times 10^{-6} \times 16 \times 10^{11}} \\
& =3 \times 10^{-5} \mathrm{~m} \\
& =0.03 \mathrm{~mm}
\end{aligned}
\)

Hint

(c) Power of the speaker: \(\boldsymbol{P}=2 \mathrm{~W}\)
Sound intensity level: \(L=120 \mathrm{~dB}\)
Reference intensity: \(I_0=10^{-12} \mathrm{~W} / \mathrm{m}^2\)

Sound intensity level formula: \(L=10 \log _{10}\left(\frac{I}{I_0}\right)\)
Intensity of sound from a point source: \(I=\frac{P}{4 \pi r^2}\)
\(
\begin{aligned}
& \text { Sound level }=10 \log _{10}\left(\frac{I}{I_0}\right) \\
& \Rightarrow 120=10 \log _{10}\left(\frac{I}{10^{-12}}\right) \\
& \Rightarrow 12=\log _{10}\left(\frac{I}{10^{-12}}\right) \\
& \Rightarrow 10^{12}=\frac{I}{10^{-12}} \\
& \Rightarrow I=1 \mathrm{~W} / \mathrm{m}^2
\end{aligned}
\)
Also we know,
\(
I=\frac{P}{4 \pi r^2}
\)
\(
\begin{aligned}
& 1=\frac{2}{4 \pi r^2} \\
& r=40 \mathrm{~cm}
\end{aligned}
\)

Hint

(b) Frequency of tuning fork: \(f=480 \mathrm{~Hz}\)
First resonance length: \(l_1=30 \mathrm{~cm}=0.3 \mathrm{~m}\)
Second resonance length: \(l_2=70 \mathrm{~cm}=0.7 \mathrm{~m}\)

The distance between two successive resonance lengths in a resonance tube experiment is equal to half the wavelength \(\left(\frac{\lambda}{2}\right)\).
The speed of sound is given by the formula \(v=f \lambda\).

How to solve
Calculate the wavelength using the difference between two successive resonance lengths and then calculate the speed of sound.

Step 1: Calculate the wavelength
The difference between two successive resonance lengths is half the wavelength:
\(l_2-l_1=\frac{\lambda}{2}\)

Substitute the given values:
\(0.7 \mathrm{~m}-0.3 \mathrm{~m}=\frac{\lambda}{2}\)

Simplify: \(0.4 \mathrm{~m}=\frac{\lambda}{2}\)

Solve for \(\lambda\) : \(\lambda=2 \times 0.4 \mathrm{~m}=0.8 \mathrm{~m}\)

Step 2: Calculate the speed of sound
Use the formula \(v=f \lambda\).
Substitute the given values:
\(v=480 \mathrm{~Hz} \times 0.8 \mathrm{~m}\)

\(v=384 \mathrm{~m} / \mathrm{s}\)
The speed of sound in air is \(384 \mathrm{~m} / \mathrm{s}\).

Hint

(b) From the question, the wave in positive \(x\)-direction is given as:
\(
Y=A \sin (k x-\omega t+\phi)
\)
Calculation:
When \(t=0, x\) becomes 0 which means \(Y=0\)
We know that, \(\sin \pi=0\)
So, \(\therefore Y=A \sin (\pi)=0\)
Therefore, the phase \(\phi=\pi\).

Hint

(c) Given, Two waves are of frequencies \(f_1=9 \mathrm{~Hz}\) and \(f_2=11 \mathrm{~Hz}\)
Beat frequency, \(\mathrm{f}_{\text {beat }}=\) difference in frequencies of two waves
\(
f_{\text {beat }}=\left|f_1-f_2\right|=11-9=2 \mathrm{~Hz}
\)
i.e., 2 Beats are produced per second
Time period of oscillation of amplitude is given by,
\(
{T}=\frac{1}{{f}_{\text {beat }}}=\frac{1}{2}=0.5 \mathrm{~s}
\)
Although the graph of oscillation is not given, the equation of envelope is given by option c.

Hint

(d) Frequency of third harmonic, \(f_3=\frac{3 v}{2 L}=240 \mathrm{~Hz}\)
\(
\begin{aligned}
& \Rightarrow \frac{3 v}{2 \times 2}=240 \mathrm{~Hz} \\
& \Rightarrow v=320 \mathrm{~ms}^{-1}
\end{aligned}
\)
Fundamental frequency, \(f_0=\frac{v}{2 L}=\frac{320}{2 \times 2}=80 \mathrm{~Hz}\)

Hint

(c) Given,
\(
Y=0.3 \sin (0.157 x) \cos (200 \pi t)
\)
So, \(k=0.157\) and \(\omega=200 \pi=2 \pi f\)
\(
\therefore \mathrm{f}=100 \mathrm{~Hz} \text { and }
\)
Speed of the sound, \(v=\frac{\omega}{k}\)
\(
=\frac{200 \pi}{0.157}=4000 \mathrm{~m} / \mathrm{s}
\)
Now, using the formula for the string or pipe closed at both the ends or open at both the ends,
\(
f=\frac{n v}{2 l}=\frac{4 v}{2 l}=\frac{2 v}{l}
\)
Here \(\mathrm{n}=4,4^{\text {th }}\) harmonic and \(l\) is the length of the string,
\(
\therefore l=\frac{2 v}{f}=\frac{2 \times 4000}{100}=80 \mathrm{~m}
\)
Thus, the length of the string is 80 m .

Hint

(c) Given, \(P=0.01 \sin [1000 t-3 x] \mathrm{Nm}^{-2}\)
On comparing with \({P}=\mathrm{P}_0 \sin (\omega {t}-{kx})\), we have \(\omega=1000 \mathrm{rad} / \mathrm{s}, {K}=3 \mathrm{~m}^{-1}\)
Speed of the sound \(=\frac{\omega}{k}\)
\(
\therefore v=\frac{\omega}{k}=\frac{1000}{3}=333.33 \mathrm{~m} / \mathrm{s}
\)
We know that,
\(
\begin{aligned}
& v \propto \sqrt{T} \\
& \frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}
\end{aligned}
\)
Or, \(\frac{333.33}{336}=\sqrt{\frac{273+0}{273+t}}\)
\(
273+t=\frac{273}{0.984}
\)
\(
\therefore \mathrm{t}=4^{\circ} \mathrm{C}
\)

Hint

(c)

Given, Length of the combined wire: \(2 L\)
Length of wire A: \(L\)
Length of wire B: \(L\)
Radius of wire A: \(r\)
Radius of wire B: \({2 r}\)
The joint of the two wires forms a node.
The linear density \(\mu\) of a wire is given by \(\mu=\rho A\), where \(\rho\) is the density and \(A\) is the cross-sectional area.
The velocity \(v\) of a wave on a string is given by \(v=\sqrt{\frac{T}{\mu}}\), where \(T\) is the tension.
The frequency \(f\) of a standing wave is given by \(f=\frac{n v}{2 L}\), where \(n\) is the number of antinodes.
How to solve?
Find the ratio of the number of antinodes by equating the frequencies of the two wires.

Step 1: Calculate the linear densities of wires A and B .
The area of wire A is \({A}_A=\pi r^2\).
The area of wire B is \(A_B=\pi(2 r)^2=4 \pi r^2\).
The linear density of wire A is \(\mu_{{A}}=\rho {A}_{{A}}=\rho \pi r^2\).
The linear density of wire B is \(\mu_B=\rho A_B=4 \rho \pi r^2\).

Step 2: Calculate the velocities of waves in wires A and B.
The velocity in wire A is \(v_A=\sqrt{\frac{T}{\mu_A}}=\sqrt{\frac{T}{\rho \pi r^2}}\).
The velocity in wire B is \(v_B=\sqrt{\frac{T}{\mu_B}}=\sqrt{\frac{T}{4 \rho \pi r^2}}=\frac{1}{2} \sqrt{\frac{T}{\rho \pi r^2}}\).

Step 3: Equate the frequencies of the two wires.
The frequency in wire A is \(f_A=\frac{p v_A}{2 L}\).
The frequency in wire B is \(f_B=\frac{q v_B}{2 L}\).
Since \(f_A=f_B\), we have \(\frac{p v_A}{2 L}=\frac{q v_B}{2 L}\).
Thus, \(p v_A=q v_B\).

Step 4: Find the ratio \(p: q\).
Substituting the expressions for \(v_A\) and \(v_B\), we get \(p \sqrt{\frac{T}{\rho \pi r^2}}=q \frac{1}{2} \sqrt{\frac{T}{\rho \pi r^2}}\).
Simplifying, we get \(p=\frac{1}{2} q\).
Therefore, the ratio \(p: q\) is \(1: 2\).

The ratio of the number of antinodes in wire A to the number of antinodes in wire \(B\) is \(1: 2\).

Hint

(b) Given, Frequency of first tuning fork: \(f_1=512 \mathrm{~Hz}\)
Length of air column for first resonance with \(f_1: I_1=11 \mathrm{~cm}=0.11 \mathrm{~m}\)
Frequency of second tuning fork: \({f}_2=256 \mathrm{~Hz}\)
Length of air column for first resonance with \(f_2: l_2=27 \mathrm{~cm}=0.27 \mathrm{~m}\)
For the first resonance in a closed tube, the length of the air column is approximately one-quarter of the wavelength: \(l \approx \frac{\lambda}{4}-e\), where \(e\) is the end correction.
The velocity of sound is given by \(v=f \lambda\).
Set up two equations using the given data and the formula for the first resonance, then solve for the velocity of sound \(v\) and the end correction e.

Set up equations for both frequencies
For \(f_1=512 \mathrm{~Hz}\) :
\(l_1=\frac{v}{4 f_1}-e\)
\(0.11=\frac{v}{4 \times 512}-e\)

For \(f_2=256 \mathrm{~Hz}\) :
\(l_2=\frac{v}{4 f_2}-e\)
\(0.27=\frac{v}{4 \times 256}-e\)

Subtract the first equation from the second equation
\(
\begin{aligned}
& (0.27-0.11)=\left(\frac{v}{4 \times 256}-e\right)-\left(\frac{v}{4 \times 512}-e\right) \\
& 0.16=\frac{v}{4 \times 256}-\frac{v}{4 \times 512} \\
& 0.16=\frac{2 v-v}{4 \times 256 \times 2} \\
& 0.16=\frac{v}{4 \times 512}
\end{aligned}
\)

Solve for \(v\)
\(
\begin{aligned}
& v=0.16 \times 4 \times 512 \\
& v=327.68 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
The velocity of sound in air is approximately \(328 \mathrm{~m} / \mathrm{s}\).

Hint

(c) How to solve?
Determine the direction of propagation from the sign in the wave equation and calculate the speed using the formula \(v=\frac{\omega}{k}\).

Step 1: Determine the direction of propagation
The given equation is \(y(x, t)=10^{-3} \sin (50 t+2 x)\).
The sign in front of \(2 x\) is positive, so the wave is propagating along the negative \(x\) axis.

Step 2: Calculate the speed of the wave
From the equation, \(\omega=50 \mathrm{rad} / \mathrm{s}\) and \(k=2 \mathrm{rad} / \mathrm{m}\).
The speed of the wave is \(v=\frac{\omega}{k}\).
\(
v=\frac{50}{2}=25 \mathrm{~m} / \mathrm{s} .
\)

The wave is propagating along the negative \(x\)-axis with a speed of \(25 \mathrm{~m} / \mathrm{s}\).

Hint

(d)

\(
\begin{aligned}
& \mathrm{y}=0.03 \sin (450 \mathrm{t}-9 \mathrm{x}) \\
& {v}=\frac{\omega}{k}=\frac{450}{9}=50 \mathrm{~m} / \mathrm{s} \\
& {v}=\sqrt{\frac{T}{\mu}} \Rightarrow \frac{T}{\mu}=2500 \\
& \Rightarrow \mathrm{~T}=2500 \times 5 \times 10^{-3} \\
& =12.5 \mathrm{~N}
\end{aligned}
\)

Hint

(c) Length of the string: \(L=1 \mathrm{~m}\)
Mass of the string: \(m=5 \mathrm{~g}=5 \times 10^{-3} \mathrm{~kg}\)
Tension in the string: \(T=8.0 \mathrm{~N}\)
Frequency of vibration: \(f=100 \mathrm{~Hz}\)

Linear mass density is given by \(\mu=\frac{m}{L}\).
Wave speed on a string is given by \(v=\sqrt{\frac{T}{\mu}}\).
The relationship between wave speed, frequency, and wavelength is \(v=f \lambda\).
The distance between successive nodes is half the wavelength, \(\frac{\lambda}{2}\).

How to solve?
Calculate the linear mass density, then the wave speed, then the wavelength, and finally the distance between successive nodes.

Step 1: Calculate the linear mass density \(\mu\)
\(
\begin{aligned}
\mu & =\frac{m}{L} \\
\mu & =\frac{5 \times 10^{-3} \mathrm{~kg}}{1 \mathrm{~m}} \\
\mu & =5 \times 10^{-3} \mathrm{~kg} / \mathrm{m}
\end{aligned}
\)

Step 2: Calculate the wave speed \(v\)
\(
\begin{aligned}
v & =\sqrt{\frac{T}{\mu}} \\
v & =\sqrt{\frac{8.0 \mathrm{~N}}{5 \times 10^{-3} \mathrm{~kg} / \mathrm{m}}} \\
v & =\sqrt{1600 \mathrm{~m}^2 / \mathrm{s}^2} \\
v & =40 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Step 3: Calculate the wavelength \(\lambda\)
\(
\begin{aligned}
v & =f \lambda \\
\lambda & =\frac{v}{f} \\
\lambda & =\frac{40 \mathrm{~m} / \mathrm{s}}{100 \mathrm{~Hz}} \\
\lambda & =0.4 \mathrm{~m}
\end{aligned}
\)

Step 4: Calculate the separation between successive nodes
Separation \(=\frac{\lambda}{2}\)
Separation \(=\frac{0.4 \mathrm{~m}}{2}\)
Separation \(=0.2 \mathrm{~m}\)
Separation \(=20 \mathrm{~cm}\)

The separation between successive nodes on the string is 20 cm.

Hint

(b)

Given: The speed of transverse waves, \(v=60 \mathrm{~ms}^{-1}\)
The wave-speed increases to \(60.5 \mathrm{~ms}^{-1}\)
To find the value of \(a\) , in terms of gravitational acceleration is closest to:
Formula of Wave speed,
\(
\Rightarrow V=\sqrt{\frac{T}{\mu}}
\)
Where, \(T\) is the tension in the string and \(\mu\) is the mass of the block per unit length ( \(l\) )
Now when car has acceleration ‘ \(a\) ‘ then the situation can be shown in fig above
Resolving the components of ‘ \(T\) ‘ along \(x\)-axis and \(y\)-axis, we get,
\(
\begin{aligned}
& \mathrm{T} \cos \theta=\mathrm{Mg} \dots(1)\\
& \mathrm{~T} \sin \theta=\mathrm{Ma} \dots(2)
\end{aligned}
\)
Squaring on both sides and adding equation (1) and (2), we get,
\(
\begin{aligned}
& \Rightarrow \mathrm{T}^2\left(\sin 2 \theta+\cos ^2 \theta\right)=\mathrm{M}^2\left(\mathrm{~g}^2+\mathrm{a}^2\right) \\
& \Rightarrow T=M\left(g^2+a^2\right)^{\frac{1}{2}} \\
& {\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]}
\end{aligned}
\)
When the car is at rest \(a=0\)
\(
\therefore 60=\sqrt{\frac{M g}{\mu}} \dots(3)
\)
Similarly, when the car is moving with acceleration a,
\(
\therefore 60.5=\sqrt{\frac{M\left(\mathrm{~g}^2+\mathrm{a}^2\right)^{1 / 2}}{\mu}} \dots(4)
\)
Dividing the equation (3) and (4) we get
\(
\begin{aligned}
& \Rightarrow \frac{60.5}{60}=\frac{\sqrt{\frac{M\left(\mathrm{~g}^2+\mathrm{a}^2\right)^{1 / 2}}{\mu}}}{\sqrt{\frac{M g}{\mu}}} \\
& \Rightarrow \frac{60.5}{60}=\sqrt{\frac{M\left(\mathrm{~g}^2+\mathrm{a}^2\right)^{1 / 2}}{\mu}} \times \frac{\mu}{M g} \\
& \Rightarrow \frac{60.5}{60}=\sqrt{\frac{\left(\mathrm{g}^2+\mathrm{a}^2\right)^{1 / 2}}{g}}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Squaring on both sides, }\\
&\begin{aligned}
& \Rightarrow\left(\frac{60.5}{60}\right)^2=\frac{\left(g^2+\mathrm{a}^2\right)^{1 / 2}}{g} \\
& \Rightarrow\left(\frac{60.5}{60}\right)^2=\sqrt{\frac{\mathrm{g}^2+\mathrm{a}^2}{g^2}} \\
& \Rightarrow\left(1+\frac{0.5}{60}\right)^2=\sqrt{1+\frac{a^2}{g^2}} \\
& \Rightarrow\left(1+\frac{0.5}{60}\right)^2=\left(1+\frac{a^2}{g^2}\right)^{1 / 2}
\end{aligned}
\end{aligned}
\)
Using binomial expansion theorem,
\(
(1+x)^n=1+\frac{n x}{1!}+\ldots \ldots
\)
On both sides, we get,
\(
\begin{aligned}
& 1+2 \times \frac{0.5}{60}=1+\frac{1}{2} \cdot \frac{a^2}{g^2} \\
& 1+\frac{1}{60}=1+\frac{1}{2} \cdot \frac{a^2}{g^2} \\
& \frac{1}{60}=\frac{1}{2} \cdot \frac{a^2}{g^2} \\
& \Rightarrow \frac{a^2}{g^2}=\frac{1}{30} \\
& \Rightarrow \frac{a}{g}=\frac{1}{\sqrt{30}} \\
& \Rightarrow a=\frac{g}{\sqrt{30}}
\end{aligned}
\)
Thus, the value of a, in terms of gravitational acceleration is closest to \(\frac{g}{5}\).

Hint

(d) The closed and open organ pipes have the same length.
The ratio of frequencies of their seventh overtones is \(\frac{a-1}{a}\).

The frequency of the \(n\)-th overtone in a closed pipe is given by \(f_c=\frac{(2 n+1) v}{4 L}\), where \(v\) is the speed of sound and \(L\) is the length of the pipe.
The frequency of the \(n\)-th overtone in an open pipe is given by \(f_o=\frac{(n+1) v}{2 L}\).

How to solve?
Calculate the frequencies of the seventh overtones for both closed and open pipes, set up the ratio, and solve for \(a\).

Step 1: Calculate the frequency of the seventh overtone for the closed pipe
For the seventh overtone, \(\boldsymbol{n}=7\).
The frequency of the seventh overtone for the closed pipe is:
\(f_c=\frac{(2(7)+1) v}{4 L}=\frac{15 v}{4 L}\)

Step 2: Calculate the frequency of the seventh overtone for the open pipe
For the seventh overtone, \(\boldsymbol{n}=7\).
The frequency of the seventh overtone for the open pipe is:
\(f_o=\frac{(7+1) v}{2 L}=\frac{8 v}{2 L}=\frac{4 v}{L}\)

Step 3: Set up the ratio of the frequencies
The ratio of the frequencies is given as \(\frac{a-1}{a}\) :
\(\frac{f_c}{f_o}=\frac{\frac{15 v}{4 L}}{\frac{4 v}{L}}=\frac{15 v}{4 L} \cdot \frac{L}{4 v}=\frac{15}{16}\)
\(\frac{a-1}{a}=\frac{15}{16}\)

Step 4: Solve for \(a\)
Cross-multiply and solve for \(a\) :
\(16(a-1)=15 a\)
\(16 a-16=15 a\)
\(16 a-15 a=16\)
\(a=16\)

The value of \(a\) is 16.

Hint

(a) Length of the first pipe: \(L_1=60 \mathrm{~cm}=0.6 \mathrm{~m}\)
Length of the second pipe: \(L_2=90 \mathrm{~cm}=0.9 \mathrm{~m}\)
Harmonic number for the first pipe: \(n_1=\mathbf{6}\)
Harmonic number for the second pipe: \(n_2=5\)
Velocity of sound in air: \(v=333 \mathrm{~m} / \mathrm{s}\)
The frequency of the \(n^{\text {th }}\) harmonic in an open organ pipe is given by \(f_n=n \frac{v}{2 L}\), where \(v\) is the velocity of sound and \(L\) is the length of the pipe.

How to solve?
Calculate the frequencies of the two harmonics and then find the difference between them.

Step 1
Calculate the frequency of the \(6^{\text {th }}\) harmonic in the first pipe.
The frequency is given by:
\(f_1=n_1 \frac{v}{2 L_1}\)
\(f_1=6 \times \frac{333}{2 \times 0.6}\)
\(f_1=\frac{1998}{1.2}\)
\(f_1=1665 \mathrm{~Hz}\)

Step 2: Calculate the frequency of the \(5^{\text {th }}\) harmonic in the second pipe.
The frequency is given by:
\(f_2=n_2 \frac{v}{2 L_2}\)
\(f_2=5 \times \frac{333}{2 \times 0.9}\)
\(f_2=\frac{1665}{1.8}\)
\(f_2=925 \mathrm{~Hz}\)

Step 3: Calculate the difference in frequencies.
The difference is given by:
\(\Delta f=\left|f_1-f_2\right|\)
\(\Delta f=|1665-925|\)
\(\Delta f=740 \mathrm{~Hz}\)

The difference in frequencies is 740 Hz.

Hint

(d) Initial resonating length \(l_1=90 \mathrm{~cm}\).
Initial fundamental frequency \(f_1=400 \mathrm{~Hz}\).
Final fundamental frequency \(f_2=600 \mathrm{~Hz}\).
Tension is constant.
The fundamental frequency of a sonometer wire is given by \(f=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}\), where \(l\) is the length, \({T}\) is the tension, and \({\mu}\) is the linear mass density.
How to solve
Use the relationship between frequency and length to find the new length.

Step 1: Set up the ratio of frequencies
Since tension and linear mass density are constant, \(\frac{f_1}{f_2}=\frac{l_2}{l_1}\).

Step 2: Solve for \(l_2\)
Rearrange the equation to solve for \(l_2: l_2=l_1 \cdot \frac{f_1}{f_2}\).

Step 3: Substitute the given values
Substitute \(l_1=90 \mathrm{~cm}, f_1=400 \mathrm{~Hz}\), and \(f_2=600 \mathrm{~Hz}\) into the equation.
\(
l_2=90 \cdot \frac{400}{600}
\)

Step 4: Calculate \(l_2\)
\(
\begin{aligned}
& l_2=90 \cdot \frac{2}{3} \\
& l_2=60 \mathrm{~cm}
\end{aligned}
\)

The resonating length of the wire with a fundamental frequency of 600 Hz is 60 cm.

Hint

(d)

\(
\begin{array}{ll}
f=\frac{1}{2} L \sqrt{\frac{T}{\mu}} \\
f_1=\frac{1}{2} \sqrt{\frac{6}{\mu}} & f_2=\frac{1}{2} \sqrt{\frac{54}{\mu}} \\
\frac{f_1}{f_2}=\frac{1}{3} & f_2-f_1=12 \\
f_1=6 H Z &
\end{array}
\)

Hint

(c) Intensity at the origin: \(I_0=16 \times 10^{-8} \mathrm{Wm}^{-2}\)
Distance 1: \(r_1=2 \mathrm{~m}\)
Distance 2: \(r_2=4 \mathrm{~m}\)

Intensity of a point source is inversely proportional to the square of the distance:
\(
I=\frac{P}{4 \pi r^2}
\)

How to solve?
Calculate the intensity at each distance and then find the difference.

Step 1: Calculate the intensity at \(r_1=2 \mathrm{~m}\)
The intensity at distance \(r_1\) is given by:
\(I_1=\frac{P}{4 \pi r_1^2}\)
We know that at the origin (very close to the source), the intensity is \(I_0=16 \times 10^{-8} \mathrm{Wm}^{-2}\).
We can express the power of the source as:
\({P}=I_0 \times 4 \pi r_{0^0}^2\), where \(r_0\) is a very small distance close to zero.
Since the intensity at the origin is given, we can assume that \(I_0\) is measured at a distance \(r_0\) such that \(I_0=\frac{P}{4 \pi r_0^2}\).
Therefore, we can write the intensity at \(r_1\) as:
\(I_1=I_0 \frac{r_0^2}{r_1^2}\)

However, since we are interested in the ratio of intensities, we can directly use the inverse square law:
\(I_1=I_0 \frac{r_0^2}{r_1^2}\)

Since we are given the intensity at the origin as \(16 \times 10^{-8} \mathrm{Wm}^{-2}\), we can consider this as the intensity at a reference distance \(r_0\).
Thus, we can write the intensity at \(r_1=2 \mathrm{~m}\) as:
\(I_1=16 \times 10^{-8} \times \frac{r_0^2}{2^2}\)

If we assume \(r_0\) is very small, we can consider the intensity at \(r_1\) relative to the intensity at a reference distance.
So, we can use the formula:
\(I_1=I_0 \frac{r_0^2}{r_1^2}\)

Since we are given the intensity at the origin, we can assume that \(I_0\) is measured at a distance \(r_0\) such that \(I_0=\frac{P}{4 \pi r_0^2}\).
Therefore, we can write the intensity at \(r_1\) as:
\(I_1=\frac{P}{4 \pi r_1^2}\)

Substituting \(P=I_0 \times 4 \pi r_0^2\), we get:
\(I_1=I_0 \frac{r_0^2}{r_1^2}\)

If we consider the intensity \(I_0\) as the intensity at a reference distance \(r_0\), we can write:
\(I_1=16 \times 10^{-8} \times \frac{r_0^2}{2^2}\)

However, since we are interested in the ratio of intensities, we can directly use the inverse square law:
\(I_1=I_0\left(\frac{r_0}{r_1}\right)^2\)
If we assume \(r_0\) is very small, we can consider the intensity at \(r_1\) relative to the intensity at a reference distance.
So, we can use the formula:
\(I_1=\frac{I_0}{r_1^2}\)

Substituting the given values:
\(I_1=\frac{16 \times 10^{-8}}{2^2}\)
\(I_1=\frac{16 \times 10^{-8}}{4}\)
\(I_1=4 \times 10^{-8} \mathrm{Wm}^{-2}\)
Step 2: Calculate the intensity at \(r_2=4 \mathrm{~m}\)
Similarly, the intensity at distance \(r_2\) is:
\(I_2=\frac{I_0}{r_2^2}\)
\(I_2=\frac{16 \times 10^{-8}}{4^2}\)
\(I_2=\frac{16 \times 10^{-8}}{16}\)
\(I_2=1 \times 10^{-8} \mathrm{Wm}^{-2}\)

Step 3: Calculate the difference in intensity
The difference in intensity is:
\(\Delta I=\left|I_1-I_2\right|\)
\(\Delta I=\left|4 \times 10^{-8}-1 \times 10^{-8}\right|\)
\(\Delta I=3 \times 10^{-8} \mathrm{Wm}^{-2}\)

The difference in intensity is \(3 \times 10^{-8} \mathrm{Wm}^{-2}\).

Hint

(d)

\(
\begin{aligned}
&\begin{aligned}
& \frac{V}{4 \ell_1}=30 \Rightarrow \ell_1=\frac{11}{4} m \\
& \frac{V}{4 \ell_2}=110 \Rightarrow \ell_2=\frac{3}{4} m \\
& \Delta \ell=2 m
\end{aligned}\\
&\text { Change in volume }=A \Delta \ell=400 \mathrm{~cm}^3\\
&M=400 \mathrm{~g} ;\left(\because \rho=1 \mathrm{~g} / \mathrm{cm}^3\right)
\end{aligned}
\)

Hint

(b)

\(
\begin{gathered}
\mathrm{f}_{\mathrm{c}}=\frac{\mathrm{v}}{4 \ell_1} \quad \mathrm{f}_{\mathrm{o}}=\frac{\mathrm{v}}{2 \ell_2} \\
\left|\mathrm{f}_{\mathrm{c}}-\mathrm{f}_0\right|=7 \\
\frac{\mathrm{v}}{4 \times 150}-\frac{\mathrm{v}}{2 \times 350}=7 \\
\frac{\mathrm{v}}{600 \mathrm{~cm}}-\frac{\mathrm{v}}{700 \mathrm{~cm}}=7 \\
\frac{\mathrm{v}}{6 \mathrm{~m}}-\frac{\mathrm{v}}{7 \mathrm{~m}}=7 \\
\mathrm{v}\left(\frac{1}{42}\right)=7 \\
\mathrm{v}=42 \times 7 \\
=294 \mathrm{~m} / \mathrm{s}
\end{gathered}
\)

Hint

(b) The fundamental frequency \(f\) is 50 Hz .
The mass of the string \(m\) is 18 g .
The linear mass density \(\mu\) is \(20 \mathrm{~g} / \mathrm{m}\).

The length of the string \(L\) can be calculated using the formula \(L=\frac{m}{\mu}\).
The speed of a transverse wave on a string is given by \(v=2 L f\), where \(L\) is the length of the string and \(f\) is the frequency.

How to solve?
First calculate the length of the string, then calculate the speed of the transverse wave.

Step 1: Calculate the length of the string

The length of the string is given by \(L=\frac{\boldsymbol{m}}{\boldsymbol{\mu}}\).
Convert the mass to \(\mathrm{kg}: m=18 \mathrm{~g}=0.018 \mathrm{~kg}\).
Convert the linear mass density to \(\mathrm{kg} / \mathrm{m}: \mu=20 \frac{\mathrm{~g}}{\mathrm{~m}}=0.020 \frac{\mathrm{~kg}}{\mathrm{~m}}\).
Calculate the length:
\(L=\frac{0.018 \mathrm{~kg}}{0.020 \frac{\mathrm{~kg}}{\mathrm{~m}}}\)
\(L=0.9 \mathrm{~m}\)

Step 2: Calculate the speed of the transverse wave
The speed of the transverse wave is given by \(v=2 L f\).
Substitute the values:
\(v=2 \cdot 0.9 \mathrm{~m} \cdot 50 \mathrm{~Hz}\)
\(v=90 \frac{\mathrm{~m}}{\mathrm{~s}}\)

The speed of the transverse wave in the string is \(90 \frac{\mathrm{~m}}{\mathrm{~s}}\).

Hint

(d) Mass \(m_1=180 \mathrm{~g}\) corresponds to frequency \(f_1=30 \mathrm{~Hz}\).
Mass \(m\) corresponds to frequency \(f_2=50 \mathrm{~Hz}\).

The frequency of a vibrating string is proportional to the square root of the tension.
Tension in the string is due to the weight of the attached mass: \({T}={m g}\).

How to solve?
Use the relationship between frequency and tension to find the unknown mass \(m\).

Step 1: Set up the ratio of frequencies and tensions
The frequency is proportional to the square root of the tension: \(f \propto \sqrt{T}\).
The tension is proportional to the mass: \({T} \propto {m}\).
Therefore, \(f \propto \sqrt{m}\).
Set up the ratio: \(\frac{f_1}{f_2}=\frac{\sqrt{m_1}}{\sqrt{m}}\).
Step 2: Solve for \(m\)
Square both sides of the equation: \(\frac{f_1^2}{f_2^2}=\frac{m_1}{m}\).
Rearrange to solve for \(m: m=m_1 \cdot \frac{f_2^2}{f_1^2}\).
Substitute the given values: \(m=180 \mathrm{~g} \cdot \frac{(50 \mathrm{~Hz})^2}{(30 \mathrm{~Hz})^2}\).
Calculate the value of \(m: m=180 \mathrm{~g} \cdot \frac{2500}{900}=180 \mathrm{~g} \cdot \frac{25}{9}=20 \mathrm{~g} \cdot 25=500 \mathrm{~g}\).

The value of \(m\) is 500 g.

Hint

(a) The first three resonance frequencies are in the ratio \(1: 3: 5\).
The frequency of the fifth harmonic is 405 Hz .
The speed of sound in air is \(324 \mathrm{~m} / \mathrm{s}\).

The resonance frequencies of a closed organ pipe are odd multiples of the fundamental frequency.
The formula for the frequency of the \(n\)-th harmonic in a closed organ pipe is \(f_n=n \frac{v}{4 L}\), where \(n\) is an odd integer, \(v\) is the speed of sound, and \(L\) is the length of the pipe.
Use the formula for the frequency of the \({n}\)-th harmonic in a closed organ pipe to find the length of the pipe.

Step 1: Find the fundamental frequency
Since the fifth harmonic corresponds to \(n=5\), we have \(f_5=5 f_1\).
Given \(f_5=405 \mathrm{~Hz}\), we can find the fundamental frequency \(f_1\) :
\(f_1=\frac{f_5}{5}=\frac{405 \mathrm{~Hz}}{5}=81 \mathrm{~Hz}\).

Step 2: Calculate the length of the pipe
Use the formula for the fundamental frequency \((n=1)\) : \(f_1=\frac{v}{4 L}\).
Rearrange the formula to solve for \(L: L=\frac{v}{4 f_1}\).
Substitute the given values: \(L=\frac{324 \mathrm{~m} / \mathrm{s}}{4 \times 81 \mathrm{~Hz}}\).
Calculate the length: \(L=\frac{324}{324} \mathrm{~m}=1 \mathrm{~m}\).

The length of the organ pipe is 1 m.

Hint

(c) In the wave equation \(\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx}+\phi)\)
Speed of wave, \(v=\frac{\omega}{k}\)
Comparing given equation,
\(
\begin{aligned}
& \omega=2 \pi(160) \\
& {K}=(0.5) 2 \pi \\
& {v}=\frac{160}{0.5}=3200 \mathrm{~m} / \mathrm{s}=320 \times \frac{18}{5} \frac{\mathrm{~km}}{\mathrm{hr}}=1152 \mathrm{~km} / \mathrm{hr}
\end{aligned}
\)

Hint

(b) \(v=\frac{w}{k}=\frac{6}{.003 \times 10^2}=\frac{6}{3}=\frac{60}{3}=20 \mathrm{~m} / \mathrm{s}\)

Hint

(d) The formula relating frequency and length is \(f=\frac{v}{2 L}\), where \(v\) is the speed of the wave on the string.
Since frequency and length are inversely proportional, we have:
\(\frac{f_1}{f_2}=\frac{L_2}{L_1}\)
\(
L_2=L_1 \cdot \frac{f_1}{f_2}=90 \mathrm{~cm} \cdot \frac{120 \mathrm{~Hz}}{180 \mathrm{~Hz}}=60 \mathrm{~cm}
\)
The length of the string producing a fundamental frequency of 180 Hz is 60 cm.

Hint

For an organ pipe open at both ends, the frequency of the \(n\)-th harmonic is given by \(f_n=n \frac{v}{2 L}\), where \(n=1,2,3, \ldots\)
Calculate the frequency of the second harmonic
Use the formula \(f_n=n \frac{v}{2 L}\) with \(n=2\).
\(
\begin{aligned}
& f_2=2 \cdot \frac{360 \mathrm{~m} / \mathrm{s}}{2 \cdot 0.4 \mathrm{~m}} \\
& f_2=\frac{720}{0.8} \mathrm{~Hz} \\
& f_2=900 \mathrm{~Hz}
\end{aligned}
\)
The frequency of the second harmonic is 900 Hz.

Hint

(a) Displacement equation of the first wave: \(y_1=10 \sin \left(\omega t+\frac{\pi}{3}\right) \mathrm{cm}\)
Displacement equation of the second wave: \(y_2=5[\sin \omega t+\sqrt{3} \cos \omega t] \mathrm{cm}\)
\(
y_2=5[\sin \omega t+\sqrt{3} \cos \omega t]
\)
\(
\begin{aligned}
& y_2=10\left[\frac{1}{2} \sin \omega t+\frac{\sqrt{3}}{2} \cos \omega t\right] \\
& y_2=10\left[\cos \left(\frac{\pi}{3}\right) \sin \omega t+\sin \left(\frac{\pi}{3}\right) \cos \omega t\right] \\
& y_2=10 \sin \left(\omega t+\frac{\pi}{3}\right)
\end{aligned}
\)
Compare the two wave equations:
\(y_1=10 \sin \left(\omega t+\frac{\pi}{3}\right)\)
\(y_2=10 \sin \left(\omega t+\frac{\pi}{3}\right)\)

The phase difference \(\phi\) is:
\(\phi=\frac{\pi}{3}-\frac{\pi}{3}=0\)
Use the formula for the resultant amplitude:
\(A=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos (\phi)}\)
\(A=\sqrt{10^2+10^2+2(10)(10) \cos (0)}\)
\(A=\sqrt{100+100+200}\)
\(A=\sqrt{400}\)
\(A=20 \mathrm{~cm}\)
The amplitude of the resultant wave is 20 cm.

Hint

(c) 

\(
\begin{aligned}
&A_R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}\\
&8=\sqrt{8^2+8^2+2 \times 8 \times 8 \cos \phi}\\
&\Rightarrow \cos \phi=-\frac{1}{2}\\
&\Rightarrow \phi=120^{\circ}
\end{aligned}
\)

Hint

(d) Frequency of the wave: \(f=500 \mathrm{~Hz}\)
Phase difference between two points: \(\Delta \phi=60^{\circ}=\frac{\pi}{3}\) rad
Distance between the two points: \(\Delta x=6.0 \mathrm{~m}\)
Relationship between phase difference, distance, and wavelength: \(\Delta \phi=\frac{2 \pi}{\lambda} \Delta x\)
Relationship between wave velocity, frequency, and wavelength: \(v=f \lambda\)
Calculate the wavelength \(\lambda\)
Use the formula: \(\Delta \phi=\frac{2 \pi}{\lambda} \Delta x\)
Rearrange to solve for \(\lambda: \lambda=\frac{2 \pi \Delta x}{\Delta \phi}\)
Substitute the given values: \(\lambda=\frac{2 \pi(6.0 \mathrm{~m})}{\frac{\pi}{3}}\)
Simplify: \(\lambda=2 \cdot 6 \cdot 3 \mathrm{~m}=36 \mathrm{~m}\)
Calculate the wave velocity \(v\)
Use the formula: \(v=f \lambda\)
Substitute the given values: \(v=(500 \mathrm{~Hz})(36 \mathrm{~m})\)
Calculate: \(v=18000 \mathrm{~m} / \mathrm{s}\)
Convert to \(\mathrm{km} / \mathrm{s}: ~ v=18 \mathrm{~km} / \mathrm{s}\)
The velocity of the wave is \(18 \mathrm{~km} / \mathrm{s}\).

Hint

(c) 

\(
\begin{aligned}
& \ell=30 \mathrm{~cm}=0.3 \mathrm{~m} \\
& \mathrm{~T}=2700 \mathrm{~N} \\
& \mathrm{f}_{\mathrm{n}}=\frac{\mathrm{nv}}{2 \ell}=\frac{\mathrm{n}}{2 \ell} \times \sqrt{\frac{\mathrm{T}}{\mu}} \\
& \mathrm{f}_{\mathrm{n}+1}=\frac{(\mathrm{n}+1)}{2 \ell} \sqrt{\frac{\mathrm{I}}{\mu}} \\
& \mathrm{f}_{\mathrm{n}}=\frac{400}{450}=\frac{\mathrm{n}}{(\mathrm{n}+1)} \\
& \mathrm{n}=8 \\
& 400=\frac{8}{2 \times 0.3} \times 30 \times \frac{\sqrt{3}}{\sqrt{\mu}} \\
& \mu=3 \mathrm{~kg} / \mathrm{m}
\end{aligned}
\)

Hint

(d) 

\(
\lambda=\frac{V}{f}=\frac{336}{400}=84 \mathrm{~cm}
\)
At first resonance, \(\frac{\lambda}{4}=l+e \Rightarrow e=1 \mathrm{~cm}\)
At \(3^{\text {rd }}\) resonance, \(l_3+e=\frac{5 \lambda}{4}\).
\(
\Rightarrow l_3=5(21)-1=104 \mathrm{~cm}
\)

Explanation:

Explanation:

\(
\begin{aligned}
&\lambda=\frac{v}{f}\\
&\text { Given that } v=336 \mathrm{~m} / \mathrm{s} \text { and } f=400 \mathrm{~Hz} \text { : }\\
&\lambda=\frac{336 \mathrm{~m} / \mathrm{s}}{400 \mathrm{~Hz}}=0.84 \mathrm{~m}=84 \mathrm{~cm}
\end{aligned}
\)

Determine the end correction:
The first resonance length is given as \(L_1=20 \mathrm{~cm}\). The effective length of the air column at the first resonance can be expressed as:
\(
L_1+e=\frac{\lambda}{4}
\)
Substituting the values:
\(
20 \mathrm{~cm}+e=\frac{84 \mathrm{~cm}}{4}=21 \mathrm{~cm}
\)
From this, we can find the end correction \(e\) :
\(
e=21 \mathrm{~cm}-20 \mathrm{~cm}=1 \mathrm{~cm}
\)
Calculate the length for the third resonance
For the third resonance, the effective length of the air column can be expressed as:
\(
L_3+e=\frac{5 \lambda}{4}
\)
Substituting the value of \({\lambda}\) :
\(
L_3+1 \mathrm{~cm}=\frac{5 \times 84 \mathrm{~cm}}{4}=105 \mathrm{~cm}
\)
Now, solving for \(L_3\) :
\(
L_3=105 \mathrm{~cm}-1 \mathrm{~cm}=104 \mathrm{~cm}
\)
The third resonance is observed when the air column has a length of 104 cm.

Hint

(a)

\(
\lambda=\frac{\mathrm{V}}{\mathrm{f}}=\frac{340}{340}=1 \mathrm{~m}
\)
First resonating length \(=\frac{\lambda}{4}=25 \mathrm{~cm}\)
Second resonating length \(=\frac{3 \lambda}{4}=75 \mathrm{~cm}\)
Third resonating length \(=\frac{5 \lambda}{4}=125 \mathrm{~cm}\)
Height of water required \(=125-75=50 \mathrm{~cm}\)

Explanation:
Frequency of tuning fork: \(f=340 \mathrm{~Hz}\)
Length of air column for first resonance: \(L_1=125 \mathrm{~cm}=1.25 \mathrm{~m}\)
Velocity of sound in air: \(v=340 \mathrm{~m} / \mathrm{s}\)
For a closed tube, resonance occurs at lengths \(L_n=\frac{(2 n-1) \lambda}{4}\), where \(n=1,2,3, \ldots\)
The first resonance (fundamental mode) occurs at \(n=1\), and the next resonance occurs at \(n=3\).
How to solve?
Calculate the wavelength of the sound wave, then find the length of the air column for the next resonance, and finally calculate the height of water required.
Step 1: Calculate the wavelength of the sound wave
Use the formula \(v=f \lambda\).
Rearrange to solve for \(\lambda: \lambda=\frac{v}{f}\).
Substitute the given values: \(\lambda=\frac{340 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~Hz}}=1 \mathrm{~m}\).

Step 2: Calculate the length of the air column for the next resonance
The next resonance occurs at \(n=3\), so \(L_3=\frac{(2(3)-1) \lambda}{4}=\frac{5 \lambda}{4}\).
Substitute the value of \(\lambda: L_3=\frac{5(1 \mathrm{~m})}{4}=1.25 \mathrm{~m}\).

Step 3: Calculate the height of water required
The height of water required is the difference between the length for the next resonance and the length for the first resonance: \(h=L_3-L_1\).
Substitute the values: \(h=1.25 \mathrm{~m}-0.25 \mathrm{~m}=1 \mathrm{~m}=100 \mathrm{~cm}-125 \mathrm{~cm}=50 \mathrm{~cm}\).

The minimum height of water required for observing resonance once again is 50 cm.

Hint

(b) There are 20 tuning forks.
Each fork gives 4 beats with the preceding fork.
The frequency of the last fork is twice the frequency of the first.

The beat frequency is the difference between the frequencies of two sound waves.
For \(n\) tuning forks with a constant beat frequency difference, the total frequency difference is \((n-1)\) times the beat frequency.

How to solve?
Calculate the total frequency difference and use the given ratio to find the frequency of the last fork.

Step 1: Calculate the total frequency difference
The total frequency difference is the number of intervals between forks multiplied by the beat frequency:
Total difference \(=(20-1) \times 4=19 \times 4=76 \mathrm{~Hz}\).

Step 2: Set up the equations
Let \(f_1\) be the frequency of the first fork and \(f_{20}\) be the frequency of the last fork. We know that \(f_{20}=2 f_1\) and \(f_{20}-f_1=76\).

Step 3: Solve for \(f_1\)
Substitute \(f_{20}=2 f_1\) into the second equation:
\(2 f_1-f_1=76\)
\(f_1=76 \mathrm{~Hz}\).

Step 4: Solve for \(f_{20}\)
Use \(f_{20}=2 f_1\) :
\(f_{20}=2 \times 76=152 \mathrm{~Hz}\).

The frequency of the last fork is 152 Hz.

Hint

(d)

\(
\begin{aligned}
& 2 \times\left(\frac{V}{2 L_0}\right)=\left(\frac{V}{4 L_c}\right) \\
& \Rightarrow L_0=4 L_c \\
& =4 \times 20 \\
& =80 \mathrm{~cm}
\end{aligned}
\)

Explanation:
The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe.
The length of the closed organ pipe is 20 cm.

The first overtone frequency of an open organ pipe is 2 times the fundamental frequency.
The fundamental frequency of an open organ pipe is \(f_{\text {open }}=\frac{v}{2 L_{\text {open }}}\).
The fundamental frequency of a closed organ pipe is \(f_{\text {closed }}=\frac{v}{4 L_{\text {closed }}}\).
How to solve
Equate the first overtone frequency of the open pipe to the fundamental frequency of the closed pipe and solve for the length of the open pipe.

Step 1: Write the equation for the first overtone frequency of the open pipe.
The first overtone frequency of an open pipe is twice its fundamental frequency:
\(
f_{\text {open_overtone }}=2 f_{\text {open }}=2 \cdot \frac{v}{2 L_{\text {open }}}=\frac{v}{L_{\text {open }}}
\)

Step 2: Write the equation for the fundamental frequency of the closed pipe.
The fundamental frequency of a closed pipe is:
\(
f_{\text {closed }}=\frac{v}{4 L_{\text {closed }}}
\)

Step 3: Equate the two frequencies and solve for \(L_{\text {open }}\).
Given that \(f_{\text {open_overtone }}=f_{\text {closed }}\) :
\(
\begin{aligned}
& \frac{v}{L_{\text {open }}}=\frac{v}{4 L_{\text {closed }}} . \\
& L_{\text {open }}=4 L_{\text {closed }} .
\end{aligned}
\)

Step 4: Substitute the given value of \(L_{\text {closed }}\) and calculate \(L_{\text {open }}\).
\(
\begin{aligned}
& L_{\text {closed }}=20 \mathrm{~cm} \\
& L_{\text {open }}=4 \cdot 20=80 \mathrm{~cm}
\end{aligned}
\)
The length of the open organ pipe is 80 cm.

Hint

(b) 

\(
\begin{aligned}
& A=|10 \cos (\pi x)| \\
& \text { At } x=\frac{4}{3} \\
& A=\left|10 \cos \left(\pi \times \frac{4}{3}\right)\right| \\
& =|-5 \mathrm{~cm}| \\
& \therefore \text { Amp }=5 \mathrm{~cm}
\end{aligned}
\)

Explanation:

The equation of the stationary wave can be expressed in the form:
\(
y=A \cos (k x) \sin (\omega t)
\)
We need to find the amplitude of the particle at \(x=\frac{4}{3} \mathrm{~cm}\). The amplitude of the particle at a specific position \(x\) in a stationary wave is given by the term \(A \cos (k x)\).
Substituting \(x=\frac{4}{3} \mathrm{~cm}\) into the cosine term:
\(
\cos (k x)=\cos \left(\pi \cdot \frac{4}{3}\right) =-\frac{1}{2}
\)

Calculate the amplitude at \(x=\frac{4}{3} \mathrm{~cm}\)
Now we can find the amplitude at this position:
\(
\begin{aligned}
\text { Amplitude }= & A \cos (k x)=10 \cdot \cos \left(\frac{4 \pi}{3}\right)=10 \\
& \cdot\left(-\frac{1}{2}\right)=-5 \mathrm{~cm}
\end{aligned}
\)
Since amplitude is a measure of displacement and is always taken as a positive quantity, we can state that the amplitude of the particle at \(x=\frac{4}{3} \mathrm{~cm}\) is: 5 cm.

Hint

(d) Linear mass density: \(\mu=9.0 \times 10^{-4} \frac{\mathrm{~kg}}{\mathrm{~m}}\)
Tension: \({T}={9 0 0} {N}\)
Resonant frequency: \(f_1=500 \mathrm{~Hz}\)
Next higher resonant frequency: \(f_2=550 \mathrm{~Hz}\)

The frequency of the \(n\)-th harmonic on a string fixed at both ends is given by \(f_n=n \frac{v}{2 L}\), where \(v\) is the wave speed and \(L\) is the length of the string.
The wave speed on a string is given by \(v=\sqrt{\frac{T}{\mu}}\).
How to solve?
Find the wave speed, then use the difference between consecutive resonant frequencies to find the length of the wire.

Step 1: Calculate the wave speed
The wave speed is given by:
\(v=\sqrt{\frac{T}{\mu}}\)
\(v=\sqrt{\frac{900}{9.0 \times 10^{-4}}}\)
\(v=\sqrt{10^6}\)
\(v=1000 \frac{\mathrm{~m}}{\mathrm{~s}}\)

Step 2: Determine the relationship between consecutive frequencies
The difference between consecutive resonant frequencies is constant and equal to \(\frac{v}{2 L}\) :
\(f_2-f_1=(n+1) \frac{v}{2 L}-n \frac{v}{2 L}\)
\(f_2-f_1=\frac{v}{2 L}\)

Step 3: Calculate the length of the wire
Substitute the given values and solve for \(L\) :
\(550-500=\frac{1000}{2 L}\)
\(50=\frac{1000}{2 L}\)
\(L=\frac{1000}{2 \times 50}\)
\(L=\frac{1000}{100}\)
\(L=10 m\)
The length of the wire is 10 m.

Hint

(c) The resonant frequency of a closed organ pipe of length \(L\) is
\(
\mathrm{f}=\frac{n v}{4 L}
\)
Here, \({n}=\) odd positive Integer
\(v=\) speed of sound in air
For \(L\) to be minimum, \({n}=1\)
\(
\begin{aligned}
& \therefore 250=\frac{v}{4 L} \Rightarrow 250=\frac{340}{4 L} \Rightarrow L=\frac{34}{4 \times 25}=0.34 \mathrm{~m} \\
& \Rightarrow {~L}=34 \mathrm{~cm}
\end{aligned}
\)

Hint

(a) 

\(
\begin{aligned}
&y_1=A_1 \sin k(x-v t)\\
&y_1=12 \sin 6.28(x-v t)\\
&\mathrm{y}_2=5 \sin 6.28({x}-{vt}+3.5)
\end{aligned}
\)
The phase difference between two waves can be calculated as
\(
\Delta \phi=\frac{2 \pi}{\lambda} \Delta x
\)
From equation of waves, the value of
\(
\Delta {x}=({x}-{vt}+3.5)-({x}-{vt})=3.5 \mathrm{~cm}
\)
We know that, wave number is \({k}=2 \pi / \lambda\).
Now, the phase difference is
\(
\Delta \phi={k} \Delta {x}=6.28 \times 3.5=2 \pi \times 3.5=7 \pi
\)
The amplitude of resulting wave can be calculated by using the following relation,
\(
A_{n e t}=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \Delta \phi}
\)
Substituting the values in above expression,
\(
\begin{aligned}
& \begin{array}{l}
{A}_{\text {net }}=\sqrt{12^2+5^2+2 \times 12 \times 5 \times \cos 7 \pi} \\
=\sqrt{144+25+120 \times(-1)} \\
{A}_{\text {net }}=\sqrt{49}=7 \mathrm{~mm}
\end{array}
\end{aligned}
\)
Thus, the amplitude of resultant wave is 7 mm.

Hint

(a)

\(
\begin{aligned}
&\text { For node }\\
&\cos \left(1.57 \mathrm{~cm}^{-1}\right) \mathrm{x}=0\\
&\left(1.57 \mathrm{~cm}^{-1}\right) \mathrm{x}=\frac{\pi}{2}\\
&\mathrm{x}=\frac{\pi}{2(1.57)} \mathrm{cm}=1 \mathrm{~cm}
\end{aligned}
\)

Explanation: 

The position of the node closest to the origin for a given standing wave equation.
What’s given in the problem
The equation of the standing wave: \(y=1.0 \mathrm{~mm} \cos \left(1.57 \mathrm{~cm}^{-1} x\right) \sin \left(78.5 \mathrm{~s}^{-1} t\right)\)

Nodes occur where the displacement is always zero.
For a standing wave of the form \({y}={A} \cos (k x) \sin (\omega t)\), nodes occur when \(\cos (k x)=0\).

How to solve?
Find the smallest positive value of \(x\) that satisfies the condition for a node.

Step 1: Find the condition for nodes.
Nodes occur when \(\cos (k x)=0\).
\(k\) is the wave number and \(x\) is the position.

Step 2: Find the general solution for \({x}\).
\(
\begin{aligned}
& \cos (k x)=0 \text { when } k x=\frac{(2 n+1) \pi}{2}, \text { where } n \text { is an integer }(n=0,1,2, \ldots) . \\
& x=\frac{(2 n+1) \pi}{2 k}
\end{aligned}
\)

Step 3: Find the position of the node closest to the origin \((x>0)\).
The closest node to the origin occurs when \(n={0}\).
\(
x=\frac{\pi}{2 k}
\)

Step 4: Substitute the given value of \(k\) and calculate \(x\).
\(
\begin{aligned}
& k=1.57 \mathrm{~cm}^{-1} \approx \frac{\pi}{2} \mathrm{~cm}^{-1} \\
& x=\frac{\pi}{2 \times \frac{\pi}{2} \mathrm{~cm}^{-1}} \\
& x=1 \mathrm{~cm}
\end{aligned}
\)
The node closest to the origin in the region \(x>0\) is at \(x=1 \mathrm{~cm}\).

Hint

(b) At time \(t=0\) :\(y=\frac{1}{1+x^2}\)
At time \(t=1\) :\(y=\frac{1}{1+(x-2)^2}\)

Determine the maximum amplitude at both times:
For \(t=0\) : The maximum amplitude occurs when \(x=0\) :
\(
y_{\max }(t=0)=\frac{1}{1+0^2}=1 \mathrm{~m}
\)
For \(t=1\) : The maximum amplitude occurs when \(x-2=0\) (i.e., \(x=2\) ):
\(
y_{\max }(t=1)=\frac{1}{1+0^2}=1 \mathrm{~m}
\)
Identify the positions of maximum amplitude:
At \(t=0\), the maximum amplitude is at \(x=0\).
At \(t=1\), the maximum amplitude is at \(x=2\).
Calculate the distance traveled by the wave:
The wave travels from \(x=0\) to \(x=2\) in 1 second.
Therefore, the distance traveled by the wave is:
\(
\Delta x=2-0=2 \mathrm{~m}
\)
Calculate the velocity of the wave:
The velocity \(v\) of the wave is given by the formula:
\(
v=\frac{\Delta x}{\Delta t}
\)
Here, \(\Delta t=1 \mathrm{~s}\) :
\(
v=\frac{2 \mathrm{~m}}{1 \mathrm{~s}}=2 \mathrm{~m} / \mathrm{s}
\)
The velocity of the wave is \(2 \mathrm{~m} / \mathrm{s}\).

Hint

(d) First overtone of open pipe \(=\frac{v_2}{L_2}\)
First overtone of closed pipe at one end \(=\frac{3 v}{4 L}\)
As per question,
\(
\begin{aligned}
& \frac{3 V}{4 L}=\frac{V_2}{L} \\
& \Rightarrow \sqrt{\frac{B}{\rho_1}} \cdot \frac{3}{4 L}=\sqrt{\frac{B}{\rho_2}} \cdot \frac{1}{L_2}\left(\because V=\sqrt{\frac{B}{\rho}}\right) \\
& \Rightarrow L_2=\frac{4 L}{3} \sqrt{\frac{\rho_1}{\rho_2}} \ldots . . \text { (i) }
\end{aligned}
\)
According to question, the length of the open pipe is
\(
\frac{x}{3} L \sqrt{\frac{\rho_1}{\rho_2}} \dots(ii)
\)
Comparing Eqs. (i) and (ii), we get
\(
x=4
\)

Hint

(d) Mass per unit length: \(\mu=0.135 \frac{\mathrm{~g}}{\mathrm{~cm}}=0.0135 \frac{\mathrm{~kg}}{\mathrm{~m}}\)
Wave equation: \(y=-0.21 \sin (x+30 t)\)
Wave speed formula: \(v=\sqrt{\frac{T}{\mu}}\), where \(T\) is tension and \(\mu\) is mass per unit length.
General form of a wave equation: \(y=A \sin (k x \pm \omega t)\), where \(\omega\) is angular frequency and \(k\) is wave number.
Wave speed can also be expressed as \(v=\frac{\omega}{k}\).

How to solve?
Calculate the wave speed using the wave equation, then use the wave speed formula to find the tension.

Step 1: Find the wave speed \(v\).
From the wave equation \(y=-0.21 \sin (x+30 t)\), we can identify \(\omega=30 \frac{\mathrm{rad}}{\mathrm{s}}\) and \(k=1 \frac{\mathrm{rad}}{\mathrm{m}}\).
\(
v=\frac{\omega}{k}30 \frac{{~m}}{{~s}}
\)

Step 2: Calculate the tension \({T}\).
Use the formula \(v=\sqrt{\frac{T}{\mu}}\) and solve for \(T\) :
\(
T=\mu v^2=12.15 \mathrm{~N}
\)

Step 3: Express the tension in the required format.
The tension is given as \(x \times 10^{-2} N\).
\(
\begin{aligned}
& 12.15=x \times 10^{-2} \\
& x=1215
\end{aligned}
\)

Step 4: Round off to the nearest integer.
\(
x \approx 1215
\)

The value of \(x\) is 1215.

Hint

(a) 

\(
\begin{aligned}
&\text { Speed of transverse wave is }\\
&\begin{aligned}
& V=\sqrt{\frac{T}{\mu}} \\
& \ln V=\frac{1}{2} \ln T-\frac{1}{2} \ln \mu \\
& \frac{\Delta V}{V}=\frac{1}{2} \frac{\Delta T}{T} \\
& =\frac{1}{2} \times 4 \\
& \Rightarrow \frac{\Delta V}{V}=2 \%
\end{aligned}
\end{aligned}
\)

Hint

(a) Length of the organ pipe: \(L=1 \mathrm{~m}\)
Density of the gas: \(\rho_{\text {gas }}=2 \rho_{\text {air }}\)
Speed of sound in air at STP: \(v_{\text {air }}=300 \frac{\mathrm{~m}}{\mathrm{~s}}\)
Speed of sound in a gas: \(v=\sqrt{\frac{B}{\rho}}\), where \(B\) is the bulk modulus and \(\rho\) is the density.
For an open organ pipe, the frequency of the \(n\)-th harmonic is given by: \(f_n=n \frac{v}{2 L}\), where \(n=1,2,3, \ldots\)

How to solve?
Calculate the speed of sound in the gas.
Calculate the fundamental frequency.
Calculate the second harmonic frequency.
Find the difference between the second harmonic and fundamental frequencies.

Step 1: Calculate the speed of sound in the gas
The bulk modulus \({B}\) remains the same since the gas is only changing in density. The speed of sound in the gas is:
\(v_{g a s}=\sqrt{\frac{B}{\rho_{g a s}}}\)
\(v_{\text {air }}=\sqrt{\frac{B}{\rho_{a i r}}}\)

Therefore:
\(\frac{v_{\text {gas }}}{v_{\text {air }}}=\sqrt{\frac{\rho_{\text {air }}}{\rho_{\text {gas }}}}=\sqrt{\frac{\rho_{\text {air }}}{2 \rho_{\text {air }}}}=\sqrt{\frac{1}{2}}\)
\(v_{g a s}=v_{a i r} \sqrt{\frac{1}{2}}=\frac{v_{a i r}}{\sqrt{2}}=\frac{300}{\sqrt{2}} \frac{\mathrm{~m}}{\mathrm{~s}}\)

Step 2: Calculate the fundamental frequency
For the fundamental frequency, \(n=1\).
\(
f_1=\frac{v_{g a s}}{2 L}=\frac{300}{2 \cdot 1 \cdot \sqrt{2}}=\frac{150}{\sqrt{2}} \mathrm{~Hz}
\)

Step 3: Calculate the second harmonic frequency
For the second harmonic, \(n=2\).
\(
f_2=2 \frac{v_{g a s}}{2 L}=2 f_1=2 \cdot \frac{150}{\sqrt{2}}=\frac{300}{\sqrt{2}} \mathrm{~Hz}
\)

Step 4: Find the difference between the second harmonic and fundamental frequencies
\(
\Delta f=f_2-f_1=\frac{300}{\sqrt{2}}-\frac{150}{\sqrt{2}}=\frac{150}{\sqrt{2}}=\frac{150 \sqrt{2}}{2}=75 \sqrt{2} \approx 106.07 \mathrm{~Hz}
\)

The frequency difference between the fundamental and second harmonic is approximately 106.07 Hz.