JEE PYQs MCQs

Hint

(b) To find the value of \(x\), we compare the frequencies of the harmonics for the closed and open organ pipes.
Step 1: Determine the frequency of the closed pipe
For a pipe closed at one end, only odd harmonics are present. The frequency of the \(n^{\text {th }}\) harmonic is:
\(
f_c=\frac{(2 n-1) v}{4 L_c}
\)
The problem specifies the fifth harmonic, so we use the term \((2 n-1)=5\) :
\(
f_5=\frac{5 v}{4 L_c}
\)
Step 2: Determine the frequency of the open pipe
For a pipe open at both ends, all harmonics are present. The frequency is given by:
\(
f_o=\frac{n v}{2 L_o}
\)
The problem specifies the first harmonic (\(n=1\)):
\(
f_1=\frac{v}{2 L_o}
\)
Step 3: Set the frequencies equal (Unison)
Since the pipes are in unison, their frequencies are equal (\(f_5=f_1\)):
\(
\frac{5 v}{4 L_c}=\frac{v}{2 L_o}
\)
Step 4: Solve for the ratio of lengths
Cancel \(v\) from both sides and rearrange to find the ratio \(L_c / L_o\) :
\(
\begin{aligned}
& \frac{5}{4 L_c}=\frac{1}{2 L_o} \\
& \frac{L_c}{L_o}=\frac{5 \times 2}{4} \\
& \frac{L_c}{L_o}=\frac{10}{4}=\frac{5}{2}
\end{aligned}
\)
Step 5: Identify the value of \(x\)
The problem defines the ratio as \(5 / x\). By comparing:
\(
\frac{5}{2}=\frac{5}{x}
\)
The value of \(x\) is 2.

Hint

(b) The volume (\(V\)) of a sphere of radius \(R\) is given by the formula :
\(
V=\frac{4}{3} \pi R^3 \dots(i)
\)
According to the problem, the volume increases by 8 times \(\left(\mathrm{V}^{\prime}=8 \mathrm{~V}\right)\) :
\(
\begin{aligned}
& 8 V=\frac{4}{3} \pi\left(R^{\prime}\right)^3 \dots(ii) \\
& \Rightarrow 8 \times \frac{4}{3} \pi R^3=\frac{4}{3} \pi\left(R^{\prime}\right)^3 \Rightarrow R^{\prime}=2 R
\end{aligned}
\)
The intensity \((\mathrm{I})\) of a point source (of power P) is the power spread over the surface area of the sphere. The formula for intensity is :
\(
I=\frac{P}{A}=\frac{P}{4 \pi R^2} \Rightarrow I \propto \frac{1}{R^2}
\)
Using the new radius \(R^{\prime}=2 R\) :
\(
\begin{aligned}
& \frac{\mathrm{I}^{\prime}}{\mathrm{I}}=\frac{\left(1 / \mathrm{R}^{\prime 2}\right)}{\left(\frac{1}{\mathrm{R}^2}\right)}=\frac{\mathrm{R}^2}{\mathrm{R}^{\prime 2}}=\frac{\mathrm{R}^2}{4 \mathrm{R}^2}=\frac{1}{4} \\
& \Rightarrow \mathrm{I}^{\prime}=\frac{\mathrm{I}}{4}
\end{aligned}
\)
Therefore, when the volume of the spherical detector increases by 8 times, the intensity decreases by 4 times.

Hint

(b) To find the length of the open organ pipe, we can use the general formula for the harmonics of a pipe open at both ends.
Step 1: Harmonic Frequency Formula
For an open organ pipe, the frequency of the \(n^{\text {th }}\) harmonic is given by:
\(
\nu_n=\frac{n v}{2 L}
\)
Where:
\(v=\) velocity of sound (\(330 \mathrm{~m} / \mathrm{s}\))
\(L=\) length of the pipe
\(n=\) harmonic number (\(1,2,3 . .\))
Step 2: Set up the Equation
The problem gives us the frequencies for the \(3^{\text {rd }}\) and \(6^{\text {th }}\) harmonics:
\(\nu_3=\frac{3 v}{2 L}\)
\(\nu_6=\frac{6 v}{2 L}\)
The difference between these frequencies is 2200 Hz :
\(
\nu_6-\nu_3=2200
\)
\(
\frac{6 v}{2 L}-\frac{3 v}{2 L}=2200
\)
Step 3: Solve for \(L\)
Combine the terms on the left side:
\(
\frac{3 v}{2 L}=2200
\)
Now, substitute the value of \(v=330 \mathrm{~m} / \mathrm{s}\) :
\(
\frac{3 \times 330}{2 L}=2200
\)
\(
L=\frac{990}{4400} \text { meters }
\)
Step 4: Convert to Millimeters
First, simplify the fraction:
\(
L=\frac{99}{440}=\frac{9}{40}=0.225 \mathrm{~m}
\)
To convert meters to millimeters, multiply by 1000 :
\(
L=0.225 \times 1000=225 \mathrm{~mm}
\)
The length of the pipe is 225 mm.

Hint

(b) To find the ratio of the times \(t_1 / t_2\), we need to calculate the speed of the pulse in each string and then determine how long it takes for each pulse to reach the junction.
Step 1: Wave Speed Formula
The speed of a transverse wave in a stretched string is determined by the tension (\(T\)) and the linear mass density \((\mu)\) :
\(
v=\sqrt{\frac{T}{\mu}}
\)
Step 2: Calculate Speeds in String A and String B
Both strings are under the same tension, \(T=500 \mathrm{~N}\).
For String A:
\(
v_A=\sqrt{\frac{500}{2 \times 10^{-4}}}=\sqrt{250 \times 10^4}=\sqrt{25 \times 10^5}
\)
To make it easier, let’s keep it as:
\(
v_A=\sqrt{\frac{500}{2 \times 10^{-4}}}=\sqrt{2.5 \times 10^6}=1581.1 \mathrm{~m} / \mathrm{s}
\)
For String B:
\(
v_B=\sqrt{\frac{500}{4 \times 10^{-4}}}=\sqrt{125 \times 10^4}=\sqrt{1.25 \times 10^6}=1118 \mathrm{~m} / \mathrm{s}
\)
Step 3: Calculate Time to Reach the Joint
Time is defined as distance divided by speed (\(t=L / v\)).
Time for pulse from C (String A) to reach the joint:
\(
t_1=\frac{L_A}{v_A}=L_A \sqrt{\frac{\mu_A}{T}}
\)
Time for pulse from D (String B) to reach the joint:
\(
t_2=\frac{L_B}{v_B}=L_B \sqrt{\frac{\mu_B}{T}}
\)
Step 4: Calculate the Ratio \(t_1 / t_2\)
Divide the two expressions:
\(
\frac{t_1}{t_2}=\frac{L_A \sqrt{\frac{\mu_A}{T}}}{L_B \sqrt{\frac{\mu_B}{T}}}
\)
Since the tension \(T\) is the same for both, it cancels out:
\(
\frac{t_1}{t_2}=\frac{L_A}{L_B} \times \sqrt{\frac{\mu_A}{\mu_B}}
\)
Now, substitute the given values \(\left(L_A=2.5, L_B=1.5, \mu_A=2 \times 10^{-4}, \mu_B=4 \times 10^{-4}\right)\) :
\(
\begin{gathered}
\frac{t_1}{t_2}=\frac{2.5}{1.5} \times \sqrt{\frac{2 \times 10^{-4}}{4 \times 10^{-4}}} \\
\frac{t_1}{t_2}=\frac{5}{3} \times \sqrt{\frac{1}{2}} \\
\frac{t_1}{t_2}=\frac{5}{3 \sqrt{2}}
\end{gathered}
\)
If we rationalize the denominator or simplify:
\(
\frac{t_1}{t_2}=\frac{5}{3 \times 1.414} \approx 1.18
\)

Hint

(b)

\(
\begin{aligned}
& A=\sqrt{2^2+4^2+2 \times 2 \times 4 \times \cos 120^{\circ}} \\
& =\sqrt{12}=2 \sqrt{3} \\
& \tan \phi=\frac{2 \sin 120^{\circ}}{4+2 \cos 120^{\circ}}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}} \\
& \phi=\frac{\pi}{6}
\end{aligned}
\)

Explanation: To find the resultant amplitude and phase of the superposed waves, we can treat the amplitudes as vectors (phasors).
Step 1: Identify the individual wave parameters
From the given equations:
Wave 1: \({A}_1={4}\), Phase \(\phi_1={0}\)
Wave 2: \(A_2=2\), Phase \(\phi_2=\frac{2 \pi}{3}\left(120^{\circ}\right)\)
The phase difference between the two waves is \(\Delta \phi=\frac{2 \pi}{3}\).
Step 2: Calculate the Resultant Amplitude (\(\boldsymbol{A}\))
The formula for the resultant amplitude of two interfering waves is:
\(
A=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos (\Delta \phi)}
\)
Substituting the values:
\(
A=\sqrt{4^2+2^2+2(4)(2) \cos \left(\frac{2 \pi}{3}\right)}
\)
\(
A=2 \sqrt{3}
\)
Step 3: Calculate the Resultant Phase (\(\boldsymbol{\theta}\))
The phase angle \(\theta\) of the resultant wave relative to the first wave is given by:
\(
\tan \theta=\frac{A_2 \sin (\Delta \phi)}{A_1+A_2 \cos (\Delta \phi)}
\)
Substituting the values:
\(
\begin{gathered}
\tan \theta=\frac{2 \sin \left(\frac{2 \pi}{3}\right)}{4+2 \cos \left(\frac{2 \pi}{3}\right)} \\
\tan \theta=\frac{2\left(\frac{\sqrt{3}}{2}\right)}{4+2\left(-\frac{1}{2}\right)} \\
\tan \theta=\frac{\sqrt{3}}{4-1}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}
\end{gathered}
\)
Since \(\tan \theta=\frac{1}{\sqrt{3}}\), the phase angle is:
\(
\theta=\frac{\pi}{6}\left(\text { or } 30^{\circ}\right)
\)
The resultant amplitude is \(2 \sqrt{3}\) and the resultant phase is \(\frac{\pi}{6}\).

Hint

(c) To find the ratio of the velocities, we need to look at how the physical dimensions of the string (radius and material) affect the wave speed.
Step 1: The Wave Speed Formula
The velocity of a transverse wave in a stretched string is:
\(
v=\sqrt{\frac{T}{\mu}}
\)
Where \(T\) is the tension and \(\mu\) is the linear mass density (mass per unit length).
Step 2: Express \(\boldsymbol{\mu}\) in terms of Radius
Since the strings have a circular cross-section and are made of the same material (meaning they have the same density \(\rho\)), we can rewrite \(\mu\) :
\(
\mu=\frac{\text { Mass }}{\text { Length }}=\frac{\text { Volume × } \rho}{L}=\frac{(A \times L) \times \rho}{L}=A \rho
\)
For a circular cross-section, the area \(A=\pi R^2\). Therefore:
\(
\mu=\pi R^2 \rho
\)
Step 3: Relate Velocity to Radius
Now, substitute \(\mu\) back into the velocity formula:
\(
v=\sqrt{\frac{T}{\pi R^2 \rho}}
\)
Since \(T, \pi\), and \(\rho\) are constant for both strings, we can see that:
\(
\begin{gathered}
v \propto \sqrt{\frac{1}{R^2}} \\
v \propto \frac{1}{R}
\end{gathered}
\)
This shows that the velocity is inversely proportional to the radius of the string.
Step 4: Calculate the Ratio
We are given:
String 1: Radius \(=R\), Velocity \(=\boldsymbol{v}_{\mathbf{1}}\)
String 2: Radius \(=R / 2\), Velocity \(=v_2\)
Using the proportionality \(v_2 / v_1=R_1 / R_2\) :
\(
\begin{gathered}
\frac{v_2}{v_1}=\frac{R}{R / 2} \\
\frac{v_2}{v_1}=2
\end{gathered}
\)
The velocity in the thinner string is twice as fast as in the thicker string.

Hint

(d) To find the minimum distance between two points having the same oscillating speed, we need to analyze the velocity of the particles in the string.
Step 1: Find the Particle Velocity
The particle velocity \(\left(v_p\right)\) is the rate of change of displacement \((y)\) with respect to time \((t)\) :
\(
v_p=\frac{\partial y}{\partial t}
\)
Given the wave equation:
\(
y=\sin (20 \pi x+10 \pi t)
\)
Differentiating with respect to \(t\) :
\(
v_p=10 \pi \cos (20 \pi x+10 \pi t)
\)
Step 2: Determine the Condition for Same Speed
“Speed” is the magnitude of velocity \(\left(\left|v_p\right|\right)\). For two points \(x_1\) and \(x_2\) to have the same speed at any time \(t\) :
\(
\left|10 \pi \cos \left(20 \pi x_1+10 \pi t\right)\right|=\left|10 \pi \cos \left(20 \pi x_2+10 \pi t\right)\right|
\)
This occurs when the arguments of the cosine function differ by a phase of \(\pi\) (since
\(
|\cos (\theta)|=|\cos (\theta+\pi)|) .
\)
Step 3: Find the Wavelength (\(\lambda\))
Compare the given equation with the standard wave equation \(y=A \sin (k x+\omega t)\) :
Wave number \(k=20 \pi\)
Angular frequency \(\omega=10 \pi\)
We know that \(k=\frac{2 \pi}{\lambda}\). Substituting the value of \(k\) :
\(
\begin{gathered}
20 \pi=\frac{2 \pi}{\lambda} \\
\lambda=\frac{2 \pi}{20 \pi}=0.1 \mathrm{~m}
\end{gathered}
\)
Step 4: Calculate the Minimum Distance
In a traveling wave, the particle speed \(\left|v_p\right|\) repeats every half-wavelength (\(\lambda / 2\)).
Think of it this way: the velocity involves a cos function. The magnitude of a cosine function repeats every \(180^{\circ}\) (or \(\pi\) radians). Since a full wavelength \(\lambda\) corresponds to a phase shift of \(2 \pi\) , a phase shift of \(\pi\) corresponds to:
\(
\text { Distance }=\frac{\lambda}{2}
\)
Substituting the value of \(\lambda\) :
\(
\text { Distance }=\frac{0.1 \mathrm{~m}}{2}=0.05 \mathrm{~m}
\)
The minimum distance is 0.05 m (or 5 cm ).

Hint

(c) To find the beat period of the superposed waves, we need to decompose the given product equation into its individual wave components using trigonometric identities.
Step 1: Use Trigonometric Identities
The given equation is in the form of a product:
\(
x=a \cos (1.5 t) \cos (50.5 t)
\)
We use the product-to-sum formula: \(\cos (A) \cos (B)=\frac{1}{2}[\cos (A+B)+\cos (A-B)]\).
Let \(A=50.5 t\) and \(B=1.5 t\).
\(
\begin{gathered}
x=\frac{a}{2}[\cos (50.5 t+1.5 t)+\cos (50.5 t-1.5 t)] \\
x=\frac{a}{2}[\cos (52 t)+\cos (49 t)]
\end{gathered}
\)
Step 2: Identify Individual Frequencies
The resulting expression shows two waves with angular frequencies \(\omega_1\) and \(\omega_2\) :
\(\omega_1=52 \mathrm{rad} / \mathrm{s}\)
\(\omega_2=49 \mathrm{rad} / \mathrm{s}\)
The frequencies in Hertz (\(f\)) are related to angular frequency by \(\omega=2 \pi f\), so:
\(
f_1=\frac{52}{2 \pi} \quad \text { and } \quad f_2=\frac{49}{2 \pi}
\)
Step 3: Calculate Beat Frequency (\(f_b\))
The beat frequency is the absolute difference between the frequencies of the two interfering waves:
\(
\begin{gathered}
f_b=\left|f_1-f_2\right| \\
f_b=\frac{52}{2 \pi}-\frac{49}{2 \pi}=\frac{3}{2 \pi} \mathrm{~Hz}
\end{gathered}
\)
Step 4: Find the Beat Period (\(T_b\))
The beat period is the reciprocal of the beat frequency:
\(
\begin{aligned}
& T_b=\frac{1}{f_b} \\
& T_b=\frac{2 \pi}{3}
\end{aligned}
\)
Now, substitute the approximate value of \(\pi \approx 3.14159\) :
\(
\begin{gathered}
T_b=\frac{2 \times 3.14159}{3} \\
T_b=\frac{6.283}{3} \approx 2.094 \text { seconds }
\end{gathered}
\)
Final Answer: Rounding to the nearest integer, the period with which they beat is 2 seconds.

Hint

(c) To find the wavelength of the wave, we need to extract the frequency from the given displacement equation and use the relationship between wave speed, frequency, and wavelength.
Step 1: Identify Angular Frequency (\(\boldsymbol{\omega}\))
The given equation for displacement is:
\(
x(t)=5 \cos \left(628 t+\frac{\pi}{2}\right)
\)
This follows the standard form \(x(t)=A \cos (\omega t+\phi)\). By comparing the two, we find the angular frequency:
\(
\omega=628 \mathrm{rad} / \mathrm{s}
\)
Step 2: Calculate the Frequency (\(f\))
We know the relationship between angular frequency and linear frequency is:
\(
\omega=2 \pi f
\)
Substituting the values (using \(\pi=3.14\)):
\(
628=2 \times 3.14 \times f
\)
\(
\begin{gathered}
628=6.28 \times f \\
f=\frac{628}{6.28}=100 \mathrm{~Hz}
\end{gathered}
\)
Step 3: Calculate the Wavelength (\(\lambda\))
The velocity of a wave \((v)\) is related to its frequency \((f)\) and wavelength \((\lambda)\) by the formula:
\(
v=f \lambda \Longrightarrow \lambda=\frac{v}{f}
\)
Given the velocity \(v=300 \mathrm{~m} / \mathrm{s}\) and our calculated frequency \(f=100 \mathrm{~Hz}\) :
\(
\begin{aligned}
& \lambda=\frac{300}{100} \\
& \lambda=3 \mathrm{~m}
\end{aligned}
\)
Final Answer: The wavelength of the wave is 3 m.

Hint

(b) To find how the frequency changes, we need to look at the relationship between the length of the air column and the fundamental frequency of a closed organ pipe.
Step 1: Fundamental Frequency of a Closed Pipe
The fundamental frequency (\(f\)) of a closed organ pipe of length \(L\) is given by:
\(
f=\frac{v}{4 L}
\)
where \(v\) is the speed of sound in air. This shows that frequency is inversely proportional to the length of the air column (\(f \propto \frac{1}{L}\)).
Step 2: Determine the New Length of the Air Column
When the pipe is filled with water, the “effective length” of the organ pipe is reduced because the air column only exists above the water level.
Initial Length: \(L_1=L\)
Water Level: \(\frac{1}{5} L\)
New Air Column Length \(\left(L_2\right): L-\frac{1}{5} L=\frac{4}{5} L\)
Step 3: Calculate the New Frequency (\(f_2\))
Using the inverse relationship:
\(
\begin{gathered}
\frac{f_2}{f_1}=\frac{L_1}{L_2} \\
\frac{f_2}{f_1}=\frac{L}{\frac{4}{5} L}=\frac{5}{4} \\
f_2=1.25 f_1
\end{gathered}
\)
Step 4: Find the Percentage Change
The fractional change in frequency is:
\(
\begin{gathered}
\text { Percentage Change }=\frac{f_2-f_1}{f_1} \times 100 \\
\text { Percentage Change }=\frac{1.25 f_1-f_1}{f_1} \times 100 \\
\text { Percentage Change }=0.25 \times 100=25 \%
\end{gathered}
\)
Since \(f_2>f_1\), the frequency increases.
Final Answer: The frequency of the fundamental note will change by \(25 \%\).

Hint

(d) To find the velocity of sound, we use the relationship between the length of a closed organ pipe and its fundamental frequency.
Step 1: Fundamental Frequency Formula
For a pipe closed at one end, the fundamental frequency \((f)\) is given by:
\(
f=\frac{v}{4 L}
\)
Where:
\(v=\) velocity of sound
\(L=\) length of the air column
Step 2: Set up the Frequencies
We have two pipes with lengths \(L_1\) and \(L_2\). Note that \(L_1<L_2\), which means \(f_1>f_2\).
\(L_1=100 \mathrm{~cm}=1.0 \mathrm{~m}\)
\(L_2=120 \mathrm{~cm}=1.2 \mathrm{~m}\)
The frequencies are:
\(
\begin{aligned}
& f_1=\frac{v}{4(1.0)}=\frac{v}{4} \\
& f_2=\frac{v}{4(1.2)}=\frac{v}{4.8}
\end{aligned}
\)
Step 3: Use the Beat Frequency
The beat frequency is the difference between the two frequencies \(\left(f_1-f_2=15\right)\) :
\(
\frac{v}{4}-\frac{v}{4.8}=15
\)
Step 4: Solve for \(v\)
Find a common denominator for the fractions:
\(
\begin{gathered}
\frac{4.8 v-4 v}{4 \times 4.8}=15 \\
\frac{0.8 v}{19.2}=15
\end{gathered}
\)
Multiply both sides by 19.2:
\(
\begin{gathered}
0.8 v=15 \times 19.2 \\
0.8 v=288
\end{gathered}
\)
Divide by 0.8 :
\(
v=\frac{288}{0.8}=360 \mathrm{~m} / \mathrm{s}
\)
Final Answer: The velocity of sound in the air column is \(360 \mathrm{~m} / \mathrm{s}\).

Hint

(a) To find the correct wave equation, we need to determine the wave parameters (amplitude, wave number, and angular frequency) and ensure the initial conditions are met.
Step 1: Identify the Parameters
Amplitude (\(A\)): The maximum displacement is given as 2 cm.
Wavelength (\(\lambda\)): Given as 7.5 cm.
Wave Number (\(k\)):
\(
k=\frac{2 \pi}{\lambda}=\frac{2 \times 3.14159}{7.5} \approx 0.838 \mathrm{rad} / \mathrm{cm}
\)
Velocity (\(v\)): The wave travels 1.2 cm in 0.3 sec.
\(
v=\frac{\text { Distance }}{\text { Time }}=\frac{1.2}{0.3}=4 \mathrm{~cm} / \mathrm{s}
\)
Angular Frequency (\(\omega\)):
\(
\omega=k \times v=0.838 \times 4 \approx 3.35 \mathrm{rad} / \mathrm{s}
\)
Step 2: Formulate the General Equation
Since the wave travels along the \(+x\) direction, the general form is \(y=A \cos (k x-\omega t+\phi)\) or \(y=A \sin (k x-\omega t+\phi)\).
Substituting our calculated values:
\(
y=2 \cos (0.838 x-3.35 t+\phi)
\)
Step 3: Check Initial Conditions
The problem states that at \(t=0\) and \(x=0\), there is a crest (maximum positive displacement).
If we use the cosine function: \(\cos (0)=1\), which gives \(y=2(1)=2 \mathrm{~cm}\). This matches the “crest” condition with a phase \(\phi=0\).
If we used a sine function, we would need a phase shift of \(\pi / 2\) to start at a crest.
Step 4: Compare with Options
Looking at the choices:
A: \(y=2 \cos (0.83 x-3.35 t)\) – Matches our \(k \approx 0.83\) and \(\omega \approx 3.35\).
D: \(y=2 \cos (3.35 x-0.83 t)-\) Swaps \(k\) and \(\omega\).

Hint

(c) Step 1: Analyze Assertion A (Speed of Sound)
The speed of sound in a medium depends on its elasticity and density. The general formula is:
\(
v=\sqrt{\frac{E}{\rho}}
\)
Where \(E\) is the modulus of elasticity and \(\rho\) is the density.
In solids, the atoms are more tightly packed and the intermolecular forces are much stronger than in gases. This high elasticity (stiffness) allows the sound energy to transfer much faster.
Speed of sound in Air: \(\approx 343 \mathrm{~m} / \mathrm{s}\)
Speed of sound in Steel: \(\approx 5960 \mathrm{~m} / \mathrm{s}\)
Assertion A is True.
Step 2: Analyze Reason R (Bulk Modulus)
The Bulk Modulus (\(B\)) measures a substance’s resistance to compression.
Solids are extremely difficult to compress because their molecules are already in close contact. Therefore, they have a very high Bulk Modulus.
Gases are easily compressed (think of squeezing a balloon). Therefore, they have a very low Bulk Modulus.
The logic in Reason R is backwards. Solids have a significantly higher Bulk Modulus than gases.
Reason R is False.
Step 3: Conclusion
Since Assertion A is a scientifically accurate statement and Reason R is a factually incorrect statement:
Assertion A: True
Reason R: False

Hint

(d) To find the velocity of the wave, we need to compare the given equation with the standard wave equation and identify the wave parameters.
Step 1: Identify Wave Parameters
The standard equation for a travelling wave is:
\(
y(x, t)=A \sin (k x \pm \omega t)
\)
From your given equation, \(y(x, t)=4.0 \sin \left[20 \times 10^{-3} x+600 t\right]\), we can identify:
Wave number (\(k\)): \(20 \times 10^{-3} \mathrm{rad} / \mathrm{mm}\)
Angular frequency \((\omega): 600 \mathrm{rad} / \mathrm{s}\)
Step 2: Determine Direction
In a wave equation, look at the signs of the \(x\) and \(t\) terms:
If the signs are opposite (e.g., \(k x-\omega t\)), the wave travels in the positive \(x\)-direction.
If the signs are the same (e.g., \(k x+\omega t\)), the wave travels in the negative \(x\)-direction.
Since both terms in your equation are positive \(\left(20 \times 10^{-3} x+600 t\right)\), the wave is travelling in the negative \(x\)-direction. This means the velocity will be negative.
Step 3: Calculate the Magnitude of Velocity
The magnitude of wave velocity \((v)\) is given by the ratio of angular frequency to wave number:
\(
v=\frac{\omega}{k}
\)
Substitute the values:
\(
\begin{gathered}
v=\frac{600 \mathrm{rad} / \mathrm{s}}{20 \times 10^{-3} \mathrm{rad} / \mathrm{mm}} \\
v=\frac{600}{0.02} \mathrm{~mm} / \mathrm{s} \\
v=30,000 \mathrm{~mm} / \mathrm{s}
\end{gathered}
\)
Step 4: Convert to Meters per Second
To convert \(\mathrm{mm} / \mathrm{s}\) to \(\mathrm{m} / \mathrm{s}\), we divide by 1000 (since \(1000 \mathrm{~mm}=1 \mathrm{~m}\)):
\(
v=\frac{30,000}{1,000} \mathrm{~m} / \mathrm{s}=30 \mathrm{~m} / \mathrm{s}
\)
Combining the magnitude (\(30 \mathrm{~m} / \mathrm{s}\)) with the direction (negative), the velocity is \(-30 \mathrm{~m} / \mathrm{s}\).

Hint

(b) To find the length of the open tube, we need to combine the formulas for harmonic frequencies with the relationship between gas density and the speed of sound.
Step 1: Speed of Sound in Different Gases
The speed of sound \((v)\) in a gas is given by the formula:
\(
v=\sqrt{\frac{B}{\rho}}
\)
Where \(B\) is the Bulk Modulus and \(\rho\) is the density. Since both gases have the same Bulk Modulus, the speeds in the two tubes are:
Closed Tube \(\left(v_1\right): v_1 \propto \frac{1}{\sqrt{\rho_1}}\)
Open Tube \(\left(v_2\right): v_2 \propto \frac{1}{\sqrt{\rho_2}}\)
The ratio of the speeds is:
\(
\frac{v_1}{v_2}=\sqrt{\frac{\rho_2}{\rho_1}}
\)
Given \(\rho_1: \rho_2=1: 16\), then \(\rho_2 / \rho_1=16\).
\(
\frac{v_1}{v_2}=\sqrt{16}=4 \Longrightarrow v_1=4 v_2
\)
Step 2: Frequency of Harmonics
Closed Organ Tube \(\left(f_c\right)\) : The frequency of the \(n^{\text {th }}\) harmonic (where \(n\) must be odd) is:
\(
f_c=\frac{n v_1}{4 L_c}
\)
For the \(9^{\text {th }}\) harmonic \((n=9)\) and \(L_c=10 \mathrm{~cm}\) :
\(
f_9=\frac{9 v_1}{4(10)}=\frac{9 v_1}{40}
\)
Open Organ Tube \(\left(f_o\right)\) : The frequency of the \(m^{\text {th }}\) harmonic is:
\(
f_o=\frac{m v_2}{2 L_o}
\)
For the \(4^{\text {th }}\) harmonic \((m=4)\) :
\(
f_4=\frac{4 v_2}{2 L_o}=\frac{2 v_2}{L_o}
\)
Step 3: Equating the Frequencies
The problem states that these frequencies are identical (\(f_9=f_4\)):
\(
\frac{9 v_1}{40}=\frac{2 v_2}{L_o}
\)
Substitute \(v_1=4 v_2\) into the equation:
\(
\frac{9\left(4 v_2\right)}{40}=\frac{2 v_2}{L_o}
\)
Cancel \(v_2\) and simplify:
\(
\begin{aligned}
& \frac{36}{40}=\frac{2}{L_o} \\
& \frac{9}{10}=\frac{2}{L_o}
\end{aligned}
\)
Step 4: Solve for \(L_o\)
\(
\begin{gathered}
9 L_o=20 \\
L_o=\frac{20}{9} \mathrm{~cm}
\end{gathered}
\)
Final Answer: The length of the open tube is \(\frac{20}{9} \mathrm{~cm}\).

Hint

(c) To find the frequency of the wave, we compare the given equation to the standard form of a plane progressive wave.
Step 1: Identify the Standard Wave Equation
The standard equation for a progressive wave is:
\(
y=A \cos \left(2 \pi f t-2 \pi \frac{x}{\lambda}\right)
\)
Or, more simply:
\(
y=A \cos \left(2 \pi\left(f t-\frac{x}{\lambda}\right)\right)
\)
Where:
\(A\) is the amplitude
\(f\) is the frequency
\(\lambda\) is the wavelength
\(t\) is time and \(x\) is distance
Step 2: Compare with the Given Equation
The equation provided is:
\(
y=2 \cos 2 \pi(330 t-x)
\)
By looking at the term inside the parentheses alongside \(t\) :
Standard form: \(f \cdot t\)
Given form: \(330 \cdot t\)
Step 3: Extract the Frequency
By direct comparison, we can see that:
\(
f=330
\)
Since the units are in SI (as implied by the ” \(m\) ” for displacement), the frequency is measured in Hertz (Hz).
Final Answer: The frequency of the wave is \(\mathbf{3 3 0 ~ H z}\).

Hint

(a) To find the length of the closed organ pipe (\(L_c\)) based on the length of the open organ pipe (\(L_o\)), we can follow these steps:
Step 1: Identify the Frequency Formulas
Open Organ Pipe (\(L_o\)): The frequencies follow the series \(f, 2 f, 3 f \ldots\) where the fundamental is \(\frac{v}{2 L_o}\). The first overtone is the second harmonic \((n=2)\) :
\(
f_o=\frac{2 v}{2 L_o}=\frac{v}{L_o}
\)
Closed Organ Pipe (\(L_c\)): The frequencies follow the series \(f, 3 f, 5 f .\). where the fundamental is \(\frac{v}{4 L_c}\). The fundamental frequency (\(n=1\)) is:
\(
f_c=\frac{v}{4 L_c}
\)
Step 2: Set the Frequencies Equal and Solve
The problem states that the fundamental frequency of the closed pipe is equal to the first overtone of the open pipe (\(f_c=f_o\)):
\(
\frac{v}{4 L_c}=\frac{v}{L_o}
\)
Cancel the velocity (\(v\)):
\(
\frac{1}{4 L_c}=\frac{1}{L_o}
\)
Cross-multiply to isolate \(L_c\) :
\(
L_o=4 L_c \Longrightarrow L_c=\frac{L_o}{4}
\)
Plug in the value for \(L_o(60 \mathrm{~cm})\) :
\(
L_c=\frac{60 \mathrm{~cm}}{4}
\)
\(
L_c=15 \mathrm{~cm}
\)

Hint

(c) To find the frequency heard by the passenger in car Q, we use the Doppler Effect formula for sound.
Step 1: Identify the Given Values
Frequency of source \(\left(f_s\right): 400 \mathrm{~Hz}\)
Velocity of sound (\(v\)): \(360 \mathrm{~m} / \mathrm{s}\)
Velocity of source (Car P, \(v_s\)): \(20 \mathrm{~m} / \mathrm{s}\) (moving away from Q)
Velocity of observer (Car Q, \(v_o\)): \(40 \mathrm{~m} / \mathrm{s}\) (moving toward P)
Step 2: Set up the Doppler Equation
The general formula for the observed frequency (\(f^{\prime}\)) is:
\(
f^{\prime}=f_s\left(\frac{v \pm v_o}{v \pm v_s}\right)
\)
Since Car Q (observer) is chasing Car P (source):
1. The observer is moving toward the source, which increases frequency (use + in numerator).
2. The source is moving away from the observer, which decreases frequency (use + in denominator).
The specific formula for this scenario is:
\(
f^{\prime}=f_s\left(\frac{v+v_o}{v+v_s}\right)
\)
Step 3: Calculate the Result
Plug in the values:
\(
f^{\prime}=400\left(\frac{360+40}{360+20}\right)
\)
Simplify the fraction:
\(
\begin{aligned}
& f^{\prime}=400\left(\frac{400}{380}\right) \\
& f^{\prime}=400\left(\frac{20}{19}\right)
\end{aligned}
\)
Perform the final division:
\(
f^{\prime}=\frac{8000}{19} \approx 421.05 \mathrm{~Hz}
\)
The frequency heard by the passenger in car Q is approximately \(\mathbf{4 2 1 ~ H z}\).

Hint

(d) We can also multiply and divide the original equation by \(\sqrt{2}\) :
\(
y=\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin \omega t+\frac{1}{\sqrt{2}} \cos \omega t\right]
\)
Using \(\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\), we get:
\(
\begin{gathered}
y=\sqrt{2}\left[\cos \left(45^{\circ}\right) \sin \omega t+\sin \left(45^{\circ}\right) \cos \omega t\right] \\
y=\sqrt{2} \sin \left(\omega t+45^{\circ}\right)
\end{gathered}
\)
Comparing this to the standard SHM equation \(y=A \sin (\omega t+\phi)\), the amplitude is clearly \(\sqrt{2}\).

Hint

(c) This is a classic conceptual “trap” in Doppler effect problems. Let’s break it down using the steps you requested.
Step 1: Identify the Given Values
Frequency of source \(\left(f_s\right): 400 \mathrm{~Hz}\)
Velocity of source (Train engine, \(v_s\)): \(10 \mathrm{~m} / \mathrm{s}\)
Velocity of observer (Passenger, \(v_o\)): \(10 \mathrm{~m} / \mathrm{s}\)
Velocity of sound (\(v\)): \(330 \mathrm{~m} / \mathrm{s}\)
Step 2: Analyze the Relative Motion
In the Doppler effect, the frequency change depends on the relative motion between the source and the observer.
1. Both the whistle (source) and the passenger (observer) are located on the same train.
2. Therefore, they are moving at the exact same velocity in the same direction.
3. The relative velocity between the source and the observer is zero.
Step 3: Calculate the Frequency
Using the Doppler Effect formula:
\(
f^{\prime}=f_s\left(\frac{v \pm v_o}{v \pm v_s}\right)
\)
Since they are moving in the same direction:
The observer is moving toward the sound source (whistle at the front) \(\Longrightarrow+v_o\)
The source is moving away from the observer \(\Longrightarrow+v_s\)
\(
\begin{gathered}
f^{\prime}=400\left(\frac{330+10}{330+10}\right) \\
f^{\prime}=400\left(\frac{340}{340}\right) \\
f^{\prime}=400 \mathrm{~Hz}
\end{gathered}
\)
The frequency heard by the passenger inside the train is \(400 \mathrm{~Hz}\)
Key takeaway: If there is no relative motion between the source and the observer (i.e., they are part of the same moving system), the observed frequency is always equal to the actual frequency.

Hint

(c) To find the speed of transverse waves in a stretched wire, we use the relationship between the tension and the linear mass density.
Step 1: Identify the Given Values
Tension (\(T\)): 70 N
Mass per unit length \((\mu): 7.0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}\)
Step 2: Identify the Speed Formula
The speed \((v)\) of a transverse wave on a string or wire is given by the formula:
\(
v=\sqrt{\frac{T}{\mu}}
\)
Step 3: Calculate the Result
Substitute the values into the formula:
\(
v=\sqrt{\frac{70}{7.0 \times 10^{-3}}}
\)
Simplify the expression inside the square root:
\(
\begin{aligned}
& v=\sqrt{\frac{70}{0.007}} \\
& v=\sqrt{10,000}
\end{aligned}
\)
Solve for \(v\) :
\(
v=100 \mathrm{~m} / \mathrm{s}
\)

Hint

(b) To find the difference in frequencies heard by the stationary observer, we calculate the apparent frequency for each train separately using the Doppler Effect.
Step 1: Identify the Given Values
Frequency of source \(\left(f_s\right): 300 \mathrm{~Hz}\)
Velocity of sound ( \(v\)): \(330 \mathrm{~m} / \mathrm{s}\)
Velocity of trains \(\left(v_s\right): 30 \mathrm{~m} / \mathrm{s}\)
Velocity of observer \(\left(v_o\right): 0 \mathrm{~m} / \mathrm{s}\) (stationary person)
Step 2: Calculate Apparent Frequencies
Train A (Moving towards the observer):
When the source moves towards a stationary observer, the frequency increases.
\(
\begin{gathered}
f_A=f_s\left(\frac{v}{v-v_s}\right) \\
f_A=300\left(\frac{330}{330-30}\right)=300\left(\frac{330}{300}\right) \\
f_A=330 \mathrm{~Hz}
\end{gathered}
\)
Train B (Moving away from the observer):
When the source moves away from a stationary observer, the frequency decreases.
\(
\begin{gathered}
f_B=f_s\left(\frac{v}{v+v_s}\right) \\
f_B=300\left(\frac{330}{330+30}\right)=300\left(\frac{330}{360}\right) \\
f_B=300\left(\frac{11}{12}\right)=275 \mathrm{~Hz}
\end{gathered}
\)
Step 3: Calculate the Difference
The difference in frequencies heard by the person is:
\(
\begin{gathered}
\Delta f=f_A-f_B \\
\Delta f=330 \mathrm{~Hz}-275 \mathrm{~Hz} \\
\Delta f=55 \mathrm{~Hz}
\end{gathered}
\)
The approximate difference of frequencies heard by the person is 55 Hz.

Hint

(d) To find the velocity of the travelling wave, we compare the given equation to the standard wave equation.
Step 1: Identify the Standard Wave Equation
A general harmonic travelling wave is represented by:
\(
y(x, t)=A \sin (k x-\omega t)
\)
Where:
\(k\) is the wave number.
\(\omega\) is the angular frequency.
Step 2: Extract the Coefficients
Comparing the standard equation to the given equation \(y(x, t)=[0.05 \sin (8 x-4 t)]\) :
Wave number \((k): 8 \mathrm{rad} / \mathrm{m}\)
Angular frequency \((\omega): 4 \mathrm{rad} / \mathrm{s}\)
Step 3: Calculate the Wave Velocity
The velocity \((v)\) of a travelling wave is the ratio of its angular frequency to its wave number:
\(
v=\frac{\omega}{k}
\)
Substitute the values:
\(
\begin{gathered}
v=\frac{4}{8} \\
v=0.5 \mathrm{~m} / \mathrm{s}
\end{gathered}
\)

Hint

(c)

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

Explanation: To find the velocity of the wave, we compare the given equation to the standard form of a progressive wave equation.
Step 1: Identify the Standard Wave Equation
The standard equation for a travelling wave is:
\(
y=A \sin (\omega t-k x)
\)
Alternatively, it can be written by factoring out the wave number (\(k\)):
\(
y=A \sin k(v t-x)
\)
Where \(v\) is the wave velocity.
Step 2: Compare the Equations
The given equation is:
\(
y=0.5 \sin \frac{2 \pi}{\lambda}(400 t-x)
\)
By comparing this to the form \(y=A \sin k(v t-x)\), we can identify the following:
Wave number \((k): \frac{2 \pi}{\lambda}\)
Velocity (\(v\)): 400
Step 3: Verify using \(\omega\) and \(k\)
If we distribute the \(\frac{2 \pi}{\lambda}\) term inside the parentheses:
\(
y=0.5 \sin \left(\frac{2 \pi \cdot 400}{\lambda} t-\frac{2 \pi}{\lambda} x\right)
\)
Now, identifying \(\omega\) and \(k\) :
\(\omega=\frac{800 \pi}{\lambda}\)
\(k=\frac{2 \pi}{\lambda}\)
The wave velocity is calculated as:
\(
\begin{gathered}
v=\frac{\omega}{k}=\frac{\frac{800 \pi}{\lambda}}{\frac{2 \pi}{\lambda}} \\
v=\frac{800 \pi}{2 \pi}=400 \mathrm{~m} / \mathrm{s}
\end{gathered}
\)

Hint

(a) To find the wavelength for which the wave velocity equals the maximum particle velocity, we need to compare the expressions for both.
Step 1: Identify the Wave Velocity
For a wave represented by \(y=A \sin (\omega t-k x)\), the wave velocity \(\left(v_w\right)\) is the speed at which the wave profile moves through the medium:
\(
v_w=\frac{\omega}{k}
\)
Step 2: Identify the Maximum Particle Velocity
The particles of the medium oscillate in Simple Harmonic Motion (SHM). Their velocity \(\left(v_p\right)\) is found by taking the partial derivative of displacement with respect to time:
\(
v_p=\frac{\partial y}{\partial t}=A \omega \cos (\omega t-k x)
\)
The maximum particle velocity (\(v_{p, \max }\)) occurs when the cosine term is 1:
\(
v_{p, \max }=A \omega
\)
Step 3: Set the Velocities Equal
According to the problem, the wave velocity is equal to the maximum particle velocity:
\(
\begin{aligned}
v_w & =v_{p, \max } \\
\frac{\omega}{k} & =A \omega
\end{aligned}
\)
Cancel the angular frequency (\(\omega\)):
\(
\frac{1}{k}=A
\)
Substitute the formula for wave number \(\left(k=\frac{2 \pi}{\lambda}\right)\) :
\(
\begin{gathered}
\frac{1}{2 \pi / \lambda}=A \\
\frac{\lambda}{2 \pi}=A
\end{gathered}
\)
Step 4: Solve for Wavelength \((\lambda)\)
From the given equation \(y=2 \sin (\omega t-k x)\), we know the amplitude \(A=2 \mathrm{~cm}\).
\(
\begin{gathered}
\frac{\lambda}{2 \pi}=2 \\
\lambda=4 \pi \mathrm{~cm}
\end{gathered}
\)

Hint

(d) Step 1: General Wave Equation
A wave traveling in the positive \(\mathbf{x}\)-axis direction is represented by:
\(
y(x, t)=A \sin (k x-\omega t)
\)
Where:
\(k\) is the wave number
\(\omega\) is the angular frequency
Step 2: Calculate the Parameters
Given:
Wavelength \((\lambda)=4.0 \mathrm{~cm}\)
Frequency \((f\) or \(v)=100 \mathrm{~Hz}\)
Step A: Find the wave number (\(k\))
\(
k=\frac{2 \pi}{\lambda}=\frac{2 \pi}{4.0}=0.5 \pi \mathrm{~cm}^{-1}
\)
Step B: Find the angular frequency (\(\omega\))
\(
\omega=2 \pi f=2 \pi(100)=200 \pi \mathrm{~s}^{-1}
\)
Let’s plug our values into the general form:
\(
y=A \sin (0.5 \pi x-200 \pi t)
\)
Now, let’s look at Option D:
\(
y=A \sin \pi\left[\left(0.5 \mathrm{~cm}^{-1}\right) x-\left(200 \mathrm{~s}^{-1}\right) t\right]
\)
If we distribute the \(\pi\) inside the brackets:
\(
y=A \sin [0.5 \pi x-200 \pi t]
\)

Hint

(b) To solve this, we need to find the expressions for both the maximum particle velocity and the wave velocity, then set up the ratio given in the problem.
Step 1: Identify the Wave Parameters
The given equation for a longitudinal wave is:
\(
x=10 \sin 2 \pi\left(n t-\frac{x}{\lambda}\right)
\)
From this equation, we can identify:
Amplitude (\(A\)): 10 cm
Angular frequency \((\omega): 2 \pi n\)
Wave number \((k): \frac{2 \pi}{\lambda}\)
Step 2: Determine Wave Velocity (\(v_w\))
The wave velocity is the speed at which the wave profile moves through the medium. It is defined as:
\(
v_w=\frac{\omega}{k}=\frac{2 \pi n}{2 \pi / \lambda}=n \lambda
\)
Step 3: Determine Maximum Particle Velocity (\(v_p\))
The particle velocity is the derivative of displacement (\(x\)) with respect to time (\(t\)). For a wave \(x=A \sin (\omega t-k x)\) :
\(
v_p=\frac{d x}{d t}=A \omega \cos (\omega t-k x)
\)
The maximum particle velocity occurs when the cosine term is 1 :
\(
\left(v_p\right)_{\max }=A \omega=10(2 \pi n)=20 \pi n
\)
Step 4: Apply the Given Condition
The problem states that the maximum particle velocity is four times the wave velocity:
\(
\left(v_p\right)_{\max }=4 \times v_w
\)
Substitute the expressions we found:
\(
20 \pi n=4 \times(n \lambda)
\)
Step 5: Solve for Wavelength (\({\lambda}\))
Now, we simplify the equation to find \(\lambda\) :
\(
20 \pi n=4 n \lambda
\)
Divide both sides by \({4} {n}\) (assuming \({n} \neq {0}\)):
\(
\begin{aligned}
& \lambda=\frac{20 \pi}{4} \\
& \lambda=5 \pi
\end{aligned}
\)

Hint

(d) To find the velocity of sound, we need to relate the given wavelengths and the beat frequency to the standard wave speed formula.
Step 1: Calculate the Beat Frequency (\(f_b\))
Beats are produced by the difference in frequencies of two sound waves. We are told that 40 beats occur in 12 seconds.
\(
f_b=\frac{\text { Number of beats }}{\text { Time }}=\frac{40}{12}=\frac{10}{3} \mathrm{~Hz}
\)
Step 2: Relate Frequency to Velocity and Wavelength
The relationship between velocity \((v)\), frequency \((f)\), and wavelength \((\lambda)\) is:
\(
v=f \lambda \Rightarrow f=\frac{v}{\lambda}
\)
We have two wavelengths:
\(\lambda_1=4.08 \mathrm{~m}\)
\(\lambda_2=4.16 \mathrm{~m}\)
Since \(\lambda_1<\lambda_2\), the corresponding frequency \(f_1\) will be greater than \(f_2\).
Step 3: Set Up the Beat Frequency Equation
The beat frequency is the difference between the two individual frequencies:
\(
\begin{aligned}
f_b & =f_1-f_2 \\
\frac{10}{3} & =\frac{v}{\lambda_1}-\frac{v}{\lambda_2}
\end{aligned}
\)
Step 4: Substitute the Values and Solve for \(v\)
Factor out the velocity (\(v\)):
\(
\frac{10}{3}=v\left(\frac{1}{4.08}-\frac{1}{4.16}\right)
\)
\(
v=707.2 \mathrm{~m} / \mathrm{s}
\)
Conclusion: The velocity of sound in the gas is \(707.2 \mathrm{~ms}^{-1}\).

Hint

(c) To understand what happens when a wave moves from one medium to another, we need to look at the fundamental relationship between wave speed, frequency, and wavelength.
Step 1: Frequency remains constant
The frequency (\(f\)) of a wave is determined solely by the source that creates it. Once the wave is traveling, the number of oscillations per second does not change when crossing a boundary. Therefore, frequency remains constant.
Step 2: Speed decreases in a denser medium
By definition, a “denser” medium (in the context of optics or acoustics) is one where the wave encounters more resistance or a higher refractive index. This causes the speed (\(v\)) of the wave to decrease.
Step 3: Wavelength must decrease
We use the wave equation to see how the wavelength \((\lambda)\) reacts:
\(
v=f \lambda
\)
If the frequency \((f)\) is constant, then the velocity \((v)\) is directly proportional to the wavelength (\(\lambda\)).
Since the speed \((v)\) decreases upon entering the denser medium, the wavelength \((\lambda)\) must also decrease to maintain the mathematical balance of the equation.
Conclusion: When entering a denser medium:
Frequency: Constant
Speed: Decreases
Wavelength: Decreases

Hint

(a) To find the percentage change in frequency, we use the Doppler Effect formula for a moving observer and a stationary source.
Step 1: Identify the Doppler Formula
When an observer moves towards a stationary source, the apparent frequency (\(f^{\prime}\)) is higher than the actual frequency (\(f\)) and is calculated as:
\(
f^{\prime}=f\left(\frac{v+v_o}{v}\right)
\)
Where:
\(v=\) velocity of sound
\(v_o=\) velocity of the observer
\(f=\) original frequency
Step 2: Plug in the Given Values
The problem states that the observer’s velocity is one-fifth the velocity of sound:
\(
v_o=\frac{1}{5} v=0.2 v
\)
Substituting this into the formula:
\(
\begin{gathered}
f^{\prime}=f\left(\frac{v+0.2 v}{v}\right) \\
f^{\prime}=f\left(\frac{1.2 v}{v}\right) \\
f^{\prime}=1.2 f
\end{gathered}
\)
Step 3: Calculate the Change in Frequency
The change in frequency (\(\Delta f\)) is the difference between the apparent frequency and the original frequency:
\(
\begin{gathered}
\Delta f=f^{\prime}-f \\
\Delta f=1.2 f-f=0.2 f
\end{gathered}
\)
Step 4: Determine the Percentage Change
To find the percentage change, we divide the change by the original frequency and multiply by 100:
\(
\begin{gathered}
\text { Percentage Change }=\left(\frac{\Delta f}{f}\right) \times 100 \\
\text { Percentage Change }=\left(\frac{0.2 f}{f}\right) \times 100 \\
\text { Percentage Change }=0.2 \times 100=20 \%
\end{gathered}
\)

Hint

(a)

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

Explanation:

To find the amplitude of the resulting wave, we need to determine the phase difference between the two individual waves and then use the formula for the resultant amplitude of interfering waves.
Step 1: Identify the Phase of Each Wave
The general form of a wave is \(y=A \sin (\phi)\), where \(\phi\) is the phase. Let’s look at the phase terms for both given equations:
Wave 1: \(y_1=5 \sin [2 \pi(x-v t)]\)
The phase \(\phi_1=2 \pi(x-v t)\)
Wave \(2: y_2=3 \sin [2 \pi(x-v t+1.5)]\)
The phase \(\phi_2=2 \pi(x-v t)+2 \pi(1.5)\)
Step 2: Calculate the Phase Difference (\({\Delta} {\phi}\))
The phase difference is the difference between \(\phi_2\) and \(\phi_1\) :
\(
\Delta \phi=\phi_2-\phi_1
\)
\(
\Delta \phi=[2 \pi(x-v t)+3 \pi]-[2 \pi(x-v t)]
\)
\(
\Delta \phi=3 \pi \text { radians }
\)
Step 3: Interpret the Phase Difference
A phase difference of \(3 \pi\) is equivalent to \(\pi\) radians (or \(180^{\circ}\)) because:
\(
3 \pi=2 \pi+\pi
\)
Since \(2 \pi\) represents a full cycle, a phase difference of \(3 \pi\) results in destructive interference. The two waves are exactly out of phase.
Step 4: Calculate the Resultant Amplitude ( \({A}_R\))
The formula for the resultant amplitude of two interfering waves with amplitudes \(A_1\) and \(A_2\) is:
\(
A_R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos (\Delta \phi)}
\)
Given:
\(A_1=5 \mathrm{~cm}\)
\(A_2=3 \mathrm{~cm}\)
\(\Delta \phi=3 \pi \Rightarrow \cos (3 \pi)=-1\)
Substitute the values:
\(
A_R=\sqrt{5^2+3^2+2(5)(3)(-1)}
\)
\(
A_R=2 \mathrm{~cm}
\)

Shortcut: For destructive interference where \(\Delta \phi=\pi, 3 \pi, 5 \pi \ldots\), the resultant amplitude is simply the difference between the two amplitudes: \(\left|A_1-A_2\right|=|5-3|=2 \mathrm{~cm}\).)
Conclusion: The amplitude of the resulting wave is 2 cm.

Hint

(c) To find the values of the amplitude \(C\) and the phase constant \(\phi\), we need to match the general solution to the specific initial conditions provided.
Step 1: Use Initial Conditions on the First Equation
The given equation is \(x(t)=A \sin \omega t+B \cos \omega t\).
At \(t=0\), position is \(x(0)\) :
\(
x(0)=A \sin (0)+B \cos (0) \Longrightarrow x(0)=B
\)
At \(t=0\), velocity is \(v(0)\) :
First, find the velocity function \(v(t)=\frac{d x}{d t}=A \omega \cos \omega t-B \omega \sin \omega t\).
\(
v(0)=A \omega \cos (0)-B \omega \sin (0) \Longrightarrow v(0)=A \omega \Longrightarrow A=\frac{v(0)}{\omega}
\)
Step 2: Relate to the Cosine Form
The displacement is also represented as \(x(t)=C \cos (\omega t-\phi)\). Using the trigonometric identity \(\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\) :
\(
\begin{gathered}
x(t)=C(\cos \omega t \cos \phi+\sin \omega t \sin \phi) \\
x(t)=(C \sin \phi) \sin \omega t+(C \cos \phi) \cos \omega t
\end{gathered}
\)
Comparing this to our initial equation \(x(t)=A \sin \omega t+B \cos \omega t\), we get:
1. \(A=C \sin \phi\)
2. \(B=C \cos \phi\)
Step 3: Calculate Amplitude \(C\)
To find \(C\), square and add the two equations above:
\(
\begin{gathered}
A^2+B^2=(C \sin \phi)^2+(C \cos \phi)^2 \\
A^2+B^2=C^2\left(\sin ^2 \phi+\cos ^2 \phi\right) \Longrightarrow C=\sqrt{A^2+B^2}
\end{gathered}
\)
Substitute the values of \(A\) and \(B\) from Step 1:
\(
C=\sqrt{\left(\frac{v(0)}{\omega}\right)^2+x(0)^2}
\)
Step 4: Calculate Phase \(\phi\)
To find \(\phi\), divide the expression for \(A\) by the expression for \(B\) :
\(
\begin{gathered}
\frac{A}{B}=\frac{C \sin \phi}{C \cos \phi}=\tan \phi \\
\tan \phi=\frac{A}{B}=\frac{v(0) / \omega}{x(0)}=\frac{v(0)}{\omega x(0)} \\
\phi=\tan ^{-1}\left(\frac{v(0)}{\omega x(0)}\right)
\end{gathered}
\)
Final Result: The parameters for the equation \(x(t)=C \cos (\omega t-\phi)\) are:
Amplitude: \(C=\sqrt{x(0)^2+\frac{v(0)^2}{\omega^2}}\)
Phase: \(\phi=\tan ^{-1}\left(\frac{v(0)}{\omega x(0)}\right)\)

Hint

(b) To find the correct mathematical expression for the wave, we need to calculate the amplitude, wave number, and angular frequency from the given data.
Step 1: Determine the Amplitude (\({A}\))
The problem states that each point moves “to and fro through a total distance of 6 cm .”
In a wave oscillation, the total distance from one extreme to the other is equal to \(2 A\).
Therefore, \(2 A=6 \mathrm{~cm} \Longrightarrow A=3 \mathrm{~cm}=0.03 \mathrm{~m}\).
Step 2: Calculate the Angular Frequency (\(\omega\))
The frequency (\(f\)) is given as 245 Hz .
\(
\begin{gathered}
\omega=2 \pi f \\
\omega=2 \pi \times 245 \approx 6.28 \times 245 \\
\omega \approx 1538.6 \mathrm{rad} / \mathrm{s} \approx 1.5 \times 10^3 \mathrm{rad} / \mathrm{s}
\end{gathered}
\)
Step 3: Calculate the Wave Number (\(k\))
The wave number is related to the velocity (\(v\)) and angular frequency (\(\omega\)) by the formula \(v= \omega / k\).
\(
\begin{gathered}
k=\frac{\omega}{v} \\
k=\frac{1538.6}{300} \approx 5.12 \mathrm{~m}^{-1}
\end{gathered}
\)
Step 4: Construct the Wave Equation
A wave traveling along the positive \(\mathbf{x}\)-axis has the form:
\(
y(x, t)=A \sin (k x-\omega t)
\)
Substituting our calculated values:
\(A=0.03\)
\(k \approx 5.1\)
\(\omega \approx 1.5 \times 10^3\)
The resulting equation is:
\(
Y(x, t)=0.03\left[\sin \left(5.1 x-1.5 \times 10^3 t\right)\right]
\)

Hint

(a) The frequency of fork \(\boldsymbol{A}\) is \(\mathbf{3 3 5 ~ H z}\).

Explanation:
Step 1: Identify initial possibilities
The beat frequency is the absolute difference between two frequencies. Since fork \(A\) produces 5 beats/s with a 340 Hz fork, its initial frequency \(f_A\) can be:
\(
f_A=340 \pm 5 \Longrightarrow f_A=345 \mathrm{~Hz} \text { or } f_A=335 \mathrm{~Hz}
\)
Step 2: Analyze the effect of filing
Filing a tuning fork removes mass from its prongs, which increases its frequency (\(f_A^{\prime}>f_A\)).
Step 3: Evaluate the two cases
Case \(1\left(f_A=345 \mathrm{~Hz}\right)\) : If the frequency increases further (e.g., to 346 Hz), the difference from 340 Hz would increase \((346-340=6\) beats \(/ \mathrm{s})\). However, the problem states the beat frequency decreased to 2 beats/ s.
Case \(2\left({f}_{{A}}{=} \mathbf{3 3 5 ~ H z}\right)\) : If the frequency increases (e.g., to \(\mathbf{3 3 8 ~ H z}\)), the difference from 340 Hz decreases (\(|338-340|=2\) beats/s). This matches the given condition.

Hint

(d) To find the reading of the water level for the first resonance, we must account for the end correction of the tube. In a resonance column experiment, the antinode of the standing wave actually forms slightly above the open end of the tube.
Step 1: Calculate the Wavelength (\({\lambda}\))
Using the relationship between speed (\(v\)), frequency (\(f\)), and wavelength:
\(
v=f \lambda \Longrightarrow \lambda=\frac{v}{f}
\)
Given:
\(v=336 \mathrm{~m} / \mathrm{s}\)
\(f=504 \mathrm{~Hz}\)
\(
\lambda=\frac{336}{504}=\frac{2}{3} \mathrm{~m} \approx 0.6667 \mathrm{~m}
\)
In centimeters: \(\lambda=66.67 \mathrm{~cm}\)
Step 2: Calculate the End Correction (\(e\))
For a cylindrical tube, the end correction is approximately 0.6 times the radius \((r)\) of the tube.
Diameter \((D)=6 \mathrm{~cm} \Longrightarrow\) Radius \((r)=3 \mathrm{~cm}\)
\(
\begin{gathered}
e=0.6 \times r \\
e=0.6 \times 3 \mathrm{~cm}=1.8 \mathrm{~cm}
\end{gathered}
\)
Step 3: Determine the Resonance Condition
For the first resonance in a tube closed at one end (the water surface), the length of the air column plus the end correction is equal to one-quarter of the wavelength:
\(
L_1+e=\frac{\lambda}{4}
\)
Where \(L_1\) is the actual reading on the scale (the distance from the top of the tube to the water level).
Step 4: Solve for the Water Level Reading (\(L_1\))
Substitute the values into the equation:
\(
L_1+1.8 \mathrm{~cm}=\frac{66.66 \mathrm{~cm}}{4}
\)
\(
L_1=14.86 \mathrm{~cm}
\)
Conclusion: The reading of the water level in the column when the first resonance occurs is approximately 14.86 cm.

Hint

(c) To determine which equation represents a travelling (progressive) wave, we look for a specific mathematical form.
Step 1: The General Rule
For a function to represent a travelling wave, the variables \(x\) and \(t\) must appear in the combined form (\(a x \pm b t\)). Mathematically, a wave is expressed as:
\(
y=f(a x \pm b t)
\)
This ensures that the wave profile maintains its shape while shifting along the \(x\)-axis as time progresses.
Step 2: Evaluating the Options
Option A: \(y=A e^x \cos (\omega t-\theta)\)
This is a stationary distribution. The amplitude (\(A e^x\)) depends on position, but the \(x\) is not linked with \(t\) inside the oscillating function. This does not travel; it simply oscillates in place with varying intensity.
Option B: \(y=A e^{-x^2}(v t+\theta)\)
While this contains \(x\) and \(t\), they are not in the form (\(x \pm v t\)). This represents a pulse whose magnitude changes globally with time rather than shifting position.
Option C: \(y=A \sin (15 x-2 t)\)
This fits the form \(f(k x-\omega t)\), where \(k=15\) and \(\omega=2\). As \(t\) increases, the value of \(x\) must also change to keep the argument of the sine function constant, which is the definition of a travelling wave.
Option D: \(y=A \sin x \cos \omega t\)

This is the standard equation for a standing wave (or stationary wave). It is a product of a function of space and a function of time, created by two travelling waves moving in opposite directions.
Option (c) is the correct representation of a travelling wave moving in the positive \(x\)-direction.

Hint

(a) Based on the Doppler Effect analysis, we can determine which graph correctly represents the frequency \((\nu)\) heard by the observer over time.
Step 1: Approach Phase \(\left(t<t_0\right)\)
As the source moves toward the point of closest approach, the component of its velocity directed toward the observer is positive. This means the observed frequency is greater than \(\nu_0\).
As it gets closer to \(t_0\), the angle between the track and the observer increases, causing the radial velocity component to decrease. Consequently, the frequency remains above \(\nu_0\) but starts to drop.
Step 2: At Closest Approach (\(t=t_0\))
At the instant \(t_0\), the source is at the minimum distance from the observer. The velocity of the source is perpendicular to the line of sight (radial velocity is zero).
The apparent frequency is exactly equal to the actual frequency: \(\nu=\nu_0\).
Step 3: Recede Phase \(\left(t>t_0\right)\)
After passing \(t_0\), the source is moving away from the observer. The radial component of velocity is now directed away, meaning the observed frequency is lower than \(\nu_0\). As the source moves further away, the frequency continues to decrease toward a lower steady value.
Step 4: Evaluate the Graphs
(a) Correct: This graph shows the frequency starting above \(\nu_0\), decreasing smoothly, passing exactly through \(\nu_0\) at \(t_0\), and ending below \(\nu_0\). This perfectly matches the physics of the situation.
(b) Incorrect: This suggests the frequency is always at or below \(\nu_0\) and reaches a minimum at \(t_0\).
(c) Incorrect: This suggests the frequency is always at or above \(\nu_0\) and reaches a maximum at \(t_0\).
(d) Incorrect: This shows a “step” change. While a step change occurs if the source passes through the observer, a source moving along a track at a finite distance creates a smooth transition as the angle changes.

Alternate:

While approaching
\(
\nu=\nu_0\left(\frac{c}{c-v \cos \theta}\right)
\)
While receding
\(
\nu=\nu_0\left(\frac{c}{c+v \cos \theta}\right)
\)

Hint

(b)

To find the speed of the car, we need to apply the Doppler Effect in two stages: first, as the sound travels from the car to the wall, and second, as the sound reflects from the wall back to the driver.
Step 1: Analyze the Two-Stage Doppler Shift
When a source is moving toward a stationary reflector and the observer is moving with the source, the situation can be broken down as follows:
Sound reaching the wall (\(f^{\prime}\)):
The wall acts as a stationary observer. The car is a source moving toward it.
\(
f^{\prime}=f_s\left(\frac{v}{v-v_c}\right)
\)
Reflected sound reaching the driver (\(f_{\text {heard }}\)):
The wall now acts as a stationary source emitting frequency \(f^{\prime}\). The driver is an observer moving toward this source.
\(
f_{\text {heard }}=f^{\prime}\left(\frac{v+v_c}{v}\right)
\)
Step 2: Combine the Equations
By substituting the expression for \(f^{\prime}\) into the second equation, we get the combined formula for a moving source and observer approaching a stationary reflector:
\(
\begin{gathered}
f_{\text {heard }}=f_s\left(\frac{v}{v-v_c}\right)\left(\frac{v+v_c}{v}\right) \\
f_{\text {heard }}=f_s\left(\frac{v+v_c}{v-v_c}\right)
\end{gathered}
\)
Where:
\(f_s=440 \mathrm{~Hz}\) (Source frequency)
\(f_{\text {heard }}=480 \mathrm{~Hz}\) (Apparent frequency)
\(v=345 \mathrm{~m} / \mathrm{s}\) (Speed of sound)
\(v_c=\)? (Speed of the car)
Step 3: Solve for the Velocity of the Car (\(v_c\))
Plug the values into the combined formula:
\(
480=440\left(\frac{345+v_c}{345-v_c}\right)
\)
\(
v_c=\frac{345}{23}=15 \mathrm{~m} / \mathrm{s}
\)
Step 4: Convert to km/hr
To convert from \(\mathrm{m} / \mathrm{s}\) to \(\mathrm{km} / \mathrm{hr}\), multiply by 3.6 (or \(\frac{18}{5}\)):
\(
v_c=15 \times \frac{18}{5}=3 \times 18=54 \mathrm{~km} / \mathrm{hr}
\)
Conclusion: The speed of the car is \(54 \mathrm{~km} / \mathrm{hr}\).

Hint

(b) To find the order of magnitude for the displacement \(s\) of air particles, we can use dimensional analysis based on the relationships provided in the problem.
Step 1: Establish the Relationship Using Dimensional Analysis
We are given that displacement (\(s\)) depends on the pressure difference (\(\Delta p\)), the density of air \((\rho)\), the speed of sound \((v)\), and the frequency \((f)\). Let’s assume the relation is:
\(
s=C \cdot \frac{\Delta p}{\rho^a v^b f^c}
\)
Since we are told to take the multiplicative constant as 1 , we set \(C=1\). Let’s check the dimensions of each variable:
\([s]=L\)
\([\Delta p]=M L^{-1} T^{-2}\)
\([\rho]=M L^{-3}\)
\([v]=L T^{-1}\)
\([f]=T^{-1}\)
By comparing the dimensions, we find the formula that balances is:
\(
s=\frac{\Delta p}{\rho v f}
\)
(Proof: \([s]=\frac{M L^{-1} T^{-2}}{\left(M L^{-3}\right)\left(L T^{-1}\right)\left(T^{-1}\right)}=\frac{M L^{-1} T^{-2}}{M L^{-2} T^{-2}}=L\). The dimensions match perfectly.)
Step 2: Plug in the Given Values
Substitute the provided values into the derived formula:
\(\Delta p \approx 10 \mathrm{~Pa}\)
\(\rho \approx 1 \mathrm{~kg} / \mathrm{m}^3\)
\(v \approx 300 \mathrm{~m} / \mathrm{s}\)
\(f \approx 1000 \mathrm{~Hz}\)
\(
s=\frac{10}{1 \times 300 \times 1000}
\)
Step 3: Calculate the Displacement (\(s\)) in Meters
\(
\begin{aligned}
& s=\frac{10}{300,000} \\
& s=\frac{1}{30,000} \mathrm{~m}
\end{aligned}
\)
Step 4: Convert to Millimeters and Compare Options
To compare with the options, convert the result from meters to millimeters (\(1 \mathrm{~m}=1000 \mathrm{~mm}\)):
\(
\begin{aligned}
& s=\frac{1}{30,000} \times 1000 \mathrm{~mm} \\
& s=\frac{1}{30} \mathrm{~mm} \approx 0.033 \mathrm{~mm}
\end{aligned}
\)
Now, evaluate the given options:
(a) 1 mm
(b) \(\frac{3}{100} \mathrm{~mm}=0.03 \mathrm{~mm}\) (Closest to 0.033 mm)
(c) 10 mm
(d) \(\frac{1}{10} \mathrm{~mm}=0.1 \mathrm{~mm}\)
Conclusion: The calculated displacement is of the order of \(\frac{3}{100} \mathrm{~mm}\).

Hint

(a) Analyzing the Wave Physics:
In a resonance tube closed at one end, resonance occurs when the length of the air column allows for a node at the water surface and an antinode at the open top. The difference between two consecutive resonance positions (\({h}_1\) and \({h}_2\)) represents exactly half a wavelength.
Using your values:
First resonance height \(\left({h}_{\mathbf{1}}\right): 17.0 \mathrm{~cm}\)
Second resonance height (\(h_2\)): 24.5 cm
Calculating Wavelength (\(\lambda\)):
The distance between two successive resonances is:
\(
\Delta h=h_2-h_1=\frac{\lambda}{2}
\)
Substituting your numbers:
\(
\begin{gathered}
7.5 \mathrm{~cm}=\frac{\lambda}{2} \\
\lambda=15 \mathrm{~cm}=0.15 \mathrm{~m}
\end{gathered}
\)
We know, \(f=\frac{v}{\lambda}\)
\(
\Rightarrow \mathrm{f}=\frac{330}{1510^{-2}}=2200 \mathrm{~Hz}
\)

Hint

(a) Step 1: Set up the Doppler Equation
For a source (the bus) moving toward a stationary reflector (the wall) and the observer (the driver) moving with that source, the relationship between the emitted frequency (\(f_s\)) and the reflected frequency (\(f_r\)) is:
\(
f_r=f_s\left(\frac{v+v_b}{v-v_b}\right)
\)
Step 2: Substitute the Given Values
\(f_s=420 \mathrm{~Hz}\)
\(f_r=490 \mathrm{~Hz}\)
\(v=330 \mathrm{~m} / \mathrm{s}\)
\(v_b=\) speed of the bus
\(
490=420\left(\frac{330+v_b}{330-v_b}\right)
\)
Step 3: Solve for \(v_b\) in \(\mathrm{m} / \mathrm{s}\)
Divide both sides by 70 to simplify:
\(
\frac{7}{6}=\frac{330+v_b}{330-v_b}
\)
Cross-multiply:
\(
\begin{gathered}
7\left(330-v_b\right)=6\left(330+v_b\right) \\
2310-7 v_b=1980+6 v_b \\
330=13 v_b \\
v_b=\frac{330}{13} \approx 25.38 \mathrm{~m} / \mathrm{s}
\end{gathered}
\)
Step 4: Convert to \(k m / h\)
To convert from \(\mathrm{m} / \mathrm{s}\) to \(\mathrm{km} / \mathrm{h}\), we multiply by 3.6 (or \(\frac{18}{5}\)):
\(
\begin{gathered}
v_b=\left(\frac{330}{13}\right) \times 3.6 \\
v_b=\frac{1188}{13} \approx 91.38 \mathrm{~km} / \mathrm{h}
\end{gathered}
\)
When rounded to the nearest whole number as per the options, we get \(91 \mathrm{~km} / \mathrm{h}\).

Hint

(b) To find the possible wavelengths, we need to translate the physical descriptions of the crests and troughs into mathematical conditions based on the wavelength \(\lambda\).
Step 1: Analyze the Distance Between Two Crests
In a transverse wave, crests are located at positions where the phase is an even multiple of \(\pi\) (specifically \(2 n \pi\)). The distance between any two crests must be an integer multiple of the wavelength.
\(
d_{\text {crest-crest }}=n \lambda=5 \mathrm{~m}
\)
where \(n=1,2,3, \ldots\)
Step 2: Analyze the Distance Between a Crest and a Trough
A trough is located exactly half a wavelength away from a crest (and then every full wavelength after that). Therefore, the distance between any crest and any trough is an odd multiple of half a wavelength:
\(
d_{\text {crest-trough }}=(2 m+1) \frac{\lambda}{2}=1.5 \mathrm{~m}
\)
where \(m=0,1,2, \ldots\)
This can be rewritten as:
\(
(2 m+1) \lambda=3 \mathrm{~m}
\)
Step 3: Solve the System of Equations
We now have two conditions that the wavelength \(\lambda\) must satisfy simultaneously:
\(\lambda=\frac{5}{n}\)
\(\lambda=\frac{3}{2 m+1}\)
By setting them equal to each other:
\(
\begin{gathered}
\frac{5}{n}=\frac{3}{2 m+1} \Longrightarrow 5(2 m+1)=3 n \\
10 m+5=3 n
\end{gathered}
\)
Step 4: Find Possible Values for \(n\) and \(m\)
Since \(n\) and \(m\) must be integers, we test values of \(m\) :
If \(m=1: 10(1)+5=15\). Then \(3 n=15 \Longrightarrow n=5\). Using \(n=5\) in condition 1: \(\lambda=\frac{5}{5}=1 \mathrm{~m}\).
If \(m=4: 10(4)+5=45\). Then \(3 n=45 \Longrightarrow n=15\). Using \(n=15\) in condition 1: \(\lambda=\frac{5}{15}=\frac{1}{3} \mathrm{~m}\).
If \(m=7: 10(7)+5=75\). Then \(3 n=75 \Longrightarrow n=25\). Using \(n=25\) in condition 1: \(\lambda=\frac{5}{25}=\frac{1}{5} \mathrm{~m}\).
Conclusion: The sequence of possible wavelengths follows the pattern \(1, \frac{1}{3}, \frac{1}{5}, \ldots\)

Hint

(a)

To find the wavelength of the wave at the top of the rope, we need to understand how the tension in the rope changes, as tension directly affects wave speed and, consequently, wavelength.
Step 1: Identify the Relationship Between Wavelength and Tension
The frequency (\(f\)) of a wave remains constant as it travels through a medium. The relationship between velocity \((v)\), frequency \((f)\), and wavelength \((\lambda)\) is \(v=f \lambda\).
For a transverse wave on a string, the velocity is given by:
\(
v=\sqrt{\frac{T}{\mu}}
\)
Where \(T\) is the tension and \(\mu\) is the mass per unit length. Since \(f\) is constant, \(\lambda \propto v\), which means:
\(
\lambda \propto \sqrt{T}
\)
Step 2: Calculate Tension at the Bottom (\(T_1\))
At the lower end of the rope, the tension is created only by the attached block.
Mass of block \((m)=2 \mathrm{~kg}\)
\(T_1=m \times g=2 g\)
The initial wavelength at this point is \(\lambda_1=6 \mathrm{~cm}\).
Step 3: Calculate Tension at the Top (\(T_2\))
At the top of the rope, the tension must support both the block and the entire weight of the rope.
Mass of rope \((M)=6 \mathrm{~kg}\)
Mass of block \((m)=2 \mathrm{~kg}\)
\(T_2=(M+m) \times g=(6+2) g=8 g\)
Step 4: Use the Proportionality to Find \({\lambda}_{\mathbf{2}}\)
Since \(\lambda\) is proportional to the square root of the tension, we can set up a ratio:
\(
\frac{\lambda_2}{\lambda_1}=\sqrt{\frac{T_2}{T_1}}
\)
Substitute the values:
\(
\begin{gathered}
\frac{\lambda_2}{6 \mathrm{~cm}}=\sqrt{\frac{8 g}{2 g}} \\
\frac{\lambda_2}{6 \mathrm{~cm}}=\sqrt{4} \\
\frac{\lambda_2}{6 \mathrm{~cm}}=2
\end{gathered}
\)
Step 5: Solve for the Final Wavelength
\(
\lambda_2=6 \mathrm{~cm} \times 2=12 \mathrm{~cm}
\)
Conclusion: The wavelength of the wavetrain when it reaches the top of the rope is 12 cm.

Hint

(a) To find the ratio of the tensions, we need to use the relationship between the fundamental frequency of a vibrating string and its tension.
Step 1: Identify the Formula for Fundamental Frequency
The fundamental frequency (\(f\)) of a stretched string is given by the formula:
\(
f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}
\)
Where:
\(L\) is the length of the string.
\(T\) is the tension in the string.
\(\mu\) is the mass per unit length (linear density).
Step 2: Establish the Proportionality
The problem states that the strings are identical and made of the same material. This means:
The length \((L)\) is the same for both.
The linear density \((\mu)\) is the same for both.
Therefore, the frequency is directly proportional to the square root of the tension:
\(
f \propto \sqrt{T} \quad \Longrightarrow \quad f^2 \propto T
\)
Step 3: Set Up the Ratio
We can write the ratio of the tensions in terms of the frequencies:
\(
\frac{T_X}{T_Z}=\left(\frac{f_X}{f_Z}\right)^2
\)
Given:
\(f_X=450 \mathrm{~Hz}\)
\(f_Z=300 \mathrm{~Hz}\)
Step 4: Calculate the Result
Substitute the values into the ratio:
\(
\frac{T_X}{T_Z}=\left(\frac{450}{300}\right)^2=2.25
\)

Hint

(b) To find the length of the wire, we can use the relationship between consecutive resonant frequencies for a string fixed at both ends.
Step 1: Calculate the Fundamental Frequency (\(f_0\))
For a string fixed at both ends, the resonant frequencies are integer multiples of the fundamental frequency (\(f_0, 2 f_0, 3 f_0, \ldots\)). The difference between any two consecutive resonant frequencies is equal to the fundamental frequency itself.
Given consecutive frequencies:
\(f_n=420 \mathrm{~Hz}\)
\(f_{n+1}=490 \mathrm{~Hz}\)
\(
\Delta f=f_{n+1}-f_n=490-420=70 \mathrm{~Hz}
\)
So, the fundamental frequency \(f_0=70 \mathrm{~Hz}\).
Step 2: Identify the Wave Speed Formula
The fundamental frequency of a string of length \(L\) is given by:
\(
f_0=\frac{v}{2 L}
\)
Where \(v\) is the velocity of the transverse wave on the string:
\(
v=\sqrt{\frac{T}{\mu}}
\)
Step 3: Calculate the Wave Velocity (\(v\))
Given:
Tension \((T)=540 \mathrm{~N}\)
Mass per unit length \((\mu)=6.0 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\)
\(
\begin{gathered}
v=\sqrt{\frac{540}{6.0 \times 10^{-3}}} \\
v=\sqrt{\frac{540}{0.006}}=\sqrt{90,000} \\
v=300 \mathrm{~m} / \mathrm{s}
\end{gathered}
\)
Step 4: Solve for Length \((L)\)
Now, substitute \(f_0\) and \(v\) back into the fundamental frequency formula:
\(
70=\frac{300}{2 L}
\)
\(
L \approx 2.14 \mathrm{~m}
\)
Conclusion: The length of the wire is approximately 2.14 meters.

Hint

(a) To find the resultant intensity, we can use the phasor addition method. Since intensity is proportional to the square of the amplitude \(\left(I \propto A^2\right)\), we first find the resultant amplitude \(A_R\).
Step 1: Identify Individual Amplitudes
Since all three waves have the same intensity \(I_0\), they all have the same amplitude \(A\). We can represent them as vectors (phasors) with magnitude \(A\) :
Wave 1: Phase \(\phi_1=0\)
Wave 2: Phase \(\phi_2=+\frac{\pi}{4}\)
Wave 3: Phase \(\phi_3=-\frac{\pi}{4}\)
Step 2: Sum the Phasors (Vector Addition)
We break each phasor into its \(x\) (horizontal) and \(y\) (vertical) components:
\(
\begin{array}{l|l|l|l}
\text { Wave } & \text { Phase } & \text { x- } \operatorname{component}(A \cos \phi) & y-\operatorname{component}(A \sin \phi) \\
\hline 1 & 0 & A \cos (0)=A & A \sin (0)=0 \\
2 & \pi / 4 & A \cos (\pi / 4)=2 A & A \sin (\pi / 4)=2 A \\
3 & -\pi / 4 & A \cos (-\pi / 4)=2 A & A \sin (-\pi / 4)=-2 A
\end{array}
\)
Total x-component (\({A}_{{x}}\)):
\(
A_x=A+\frac{A}{\sqrt{2}}+\frac{A}{\sqrt{2}}=A+\frac{2 A}{\sqrt{2}}=A+A \sqrt{2}=A(1+\sqrt{2})
\)
Total y-component (\({A}_{{y}}\)):
\(
A_y=0+\frac{A}{\sqrt{2}}-\frac{A}{\sqrt{2}}=0
\)
Step 3: Calculate Resultant Amplitude and Intensity
The resultant amplitude \(A_R\) is:
\(
A_R=\sqrt{A_x^2+A_y^2}=A(1+\sqrt{2})
\)
Since Intensity \(I \propto A^2\), the resultant intensity \(I_R\) is:
\(
I_R=I_0(1+\sqrt{2})^2
\)
Thus, \(I_R \approx 5.8 I_0\).

Hint

(d) To find the new tension \(T\), we use the physical relationship between the tension in a string and the velocity of a transverse wave traveling through it.
Step 1: Identify the Governing Formula
The velocity (\(v\)) of a transverse wave on a taut wire is given by the formula:
\(
v=\sqrt{\frac{\text { Tension }}{\text { Linear Mass Density }}}=\sqrt{\frac{T}{\mu}}
\)
Step 2: Establish the Proportionality
Since the wire remains the same, the mass per unit length \((\mu)\) is constant. Therefore, the velocity is directly proportional to the square root of the tension:
\(
v \propto \sqrt{T}
\)
This can be rewritten as:
\(
T \propto v^2
\)
Step 3: Set Up the Ratio
Let \(v_1\) and \(T_1\) be the initial velocity and tension, and \(v_2\) and \(T_2\) be the final velocity and tension.
\(
\frac{T_2}{T_1}=\left(\frac{v_2}{v_1}\right)^2
\)
Given:
\(T_1=2.06 \times 10^4 \mathrm{~N}\)
\(v_1=v\)
\(v_2=v / 2\)
\(T_2=T\)
Step 4: Solve for \(T\)
Substitute the values into the ratio:
\(
\frac{T}{2.06 \times 10^4}=\left(\frac{v / 2}{v}\right)^2
\)
\(
T=5.15 \times 10^3 \mathrm{~N}
\)
Conclusion: The value of the new tension \(T\) is \(5.15 \times 10^3 \mathrm{~N}\).

Hint

(c) To find the speed of the tuning forks, we need to calculate the apparent frequencies heard by the stationary observer from both the approaching and receding sources, then use the beat frequency.
Step 1: Identify the Doppler Shift Formulas
Since the observer is stationary (\(v_o=0\)), the apparent frequency (\(f^{\prime}\)) depends on the speed of the source (\(v_s\)).
For the approaching fork \(\left(f_1\right)\) :
\(
f_1=f_0\left(\frac{v}{v-v_s}\right)
\)
For the receding fork (\(f_2\)):
\(
f_2=f_0\left(\frac{v}{v+v_s}\right)
\)
Step 2: Calculate the Beat Frequency (\(f_b\))
The beat frequency is the difference between these two frequencies:
\(
f_b=f_1-f_2
\)
\(
f_b=f_0\left(\frac{v}{v-v_s}-\frac{v}{v+v_s}\right)
\)
Simplify the term in the parentheses:
\(
\begin{gathered}
f_b=f_0 \cdot v\left(\frac{\left(v+v_s\right)-\left(v-v_s\right)}{\left(v-v_s\right)\left(v+v_s\right)}\right) \\
f_b=f_0 \cdot v\left(\frac{2 v_s}{v^2-v_s^2}\right)
\end{gathered}
\)
Step 3: Apply the Approximation
The problem states that \(v_s\) is much less than the speed of sound \(\left(v_s \ll v\right)\). Therefore, we can approximate \(v^2-v_s^2 \approx v^2\).
\(
\begin{gathered}
f_b \approx f_0 \cdot v\left(\frac{2 v_s}{v^2}\right) \\
f_b \approx \frac{2 f_0 v_s}{v}
\end{gathered}
\)
Step 4: Solve for the Speed of the Fork (\(v_s\))
Substitute the given values:
\(f_b=2 \mathrm{~Hz}\)
\(f_0=1400 \mathrm{~Hz}\)
\(v=350 \mathrm{~m} / \mathrm{s}\)
\(
2=\frac{2 \times 1400 \times v_s}{350}
\)
Simplify the right side:
\(
\begin{gathered}
2=\frac{2800}{350} \times v_s \\
2=8 \times v_s \\
v_s=\frac{2}{8}=\frac{1}{4} \mathrm{~m} / \mathrm{s}
\end{gathered}
\)
Conclusion: The speed of each tuning fork is \(1 / 4 \mathrm{~m} / \mathrm{s}\).

Hint

(a) To find the extension of the wire, we need to bridge the gap between wave mechanics (to find the tension) and elasticity (to find the extension).
Step 1: Calculate the Mass per Unit Length \(\boldsymbol{(} \mu \boldsymbol{)}\)
First, we find the linear mass density of the wire:
Mass \((m)=6.0 \mathrm{~g}=6 \times 10^{-3} \mathrm{~kg}\)
Length \((L)=60 \mathrm{~cm}=0.6 \mathrm{~m}\)
\(
\mu=\frac{m}{L}=\frac{6 \times 10^{-3}}{0.6}=0.01 \mathrm{~kg} / \mathrm{m}
\)
Step 2: Calculate the Tension (\(T\))
The speed of a transverse wave is related to the tension by the formula \(v=\sqrt{T / \mu}\). We can rearrange this to solve for \(T\) :
\(
T=\mu v^2
\)
Given the speed \(v=90 \mathrm{~m} / \mathrm{s}\) :
\(
T=(0.01) \times(90)^2=0.01 \times 8100=81 \mathrm{~N}
\)
Step 3: Relate Tension to Young’s Modulus
Young’s Modulus (\(Y\)) relates the stress on a material to its strain:
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T / A}{\Delta L / L}
\)
Rearranging to solve for the extension (\({\Delta} {L}\)):
\(
\Delta L=\frac{T \cdot L}{A \cdot Y}
\)
Step 4: Substitute the Values
Identify the remaining parameters in SI units:
Tension \((T)=81 \mathrm{~N}\)
Length \((L)=0.6 \mathrm{~m}\)
\(\operatorname{Area}(A)=1.0 \mathrm{~mm}^2=1.0 \times 10^{-6} \mathrm{~m}^2\)
Young’s Modulus \((Y)=16 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
\(
\Delta L=\frac{81 \times 0.6}{\left(1.0 \times 10^{-6}\right) \times\left(16 \times 10^{11}\right)}
\)
\(
\begin{gathered}
\Delta L=\frac{48.6}{16 \times 10^5} \\
\Delta L=3.0375 \times 10^{-5} \mathrm{~m}
\end{gathered}
\)
Step 5: Final Result
Convert the result to a more readable unit (millimeters):
\(
\Delta L=0.030375 \mathrm{~mm}
\)
Conclusion: The extension of the wire over its natural length is approximately 0.03 mm.