Past NEET Papers

Hint

Length of string of musical instrument,
\(
l=90 \mathrm{~cm}=0.9 \mathrm{~m}
\)
Fundamental frequency,\(f_1=120 \mathrm{~Hz}\)
\(
\begin{aligned}
& f_2=180 \mathrm{~Hz} \\
& \therefore \text { We know that } f \propto \frac{1}{l} \\
& \Rightarrow \quad \frac{f_1}{f_2}=\frac{l_2}{l_1} \Rightarrow l_2=\frac{f_1 l_1}{f_2}=\frac{120 \times 0.9}{180} \\
&=\frac{2}{3} \times 0.9=0.6 \mathrm{~m}=60 \mathrm{~cm}
\end{aligned}
\)

Hint

Frequency of string, \(f=\frac{1}{2 l} \sqrt{\frac{T}{m}}\)
Frequency \(\propto \sqrt{\text { Tension }}\)
Difference of \(f_A\) and \(f_B\) is \(6 \mathrm{~Hz}\).
In case of beats formation unknown frequency \(\left(f_B\right)=f_A \pm\) beats where, \(f_A=530 \mathrm{~Hz}\) and beats \(=6 \mathrm{~Hz}\). \(\Rightarrow f_B=530 \pm 6=536\) or \(524 \mathrm{~Hz}\) When tension in \(B\) is slightly decreased, then beats frequency increases to \(7 \mathrm{~Hz}\). This is possible if we take original frequency of \(B\) as \(524 \mathrm{~Hz}\).

Hint

\(
\begin{aligned}
& l_1=9.75 \mathrm{~cm} \\
& l_2=31.25 \mathrm{~cm} \\
& l_3=52.75 \mathrm{~cm} \\
& e=\text { end correction } \\
& \frac{\lambda}{4}+e=9.75 \mathrm{~cm} \\
& \frac{3 \lambda}{4}+e=31.25 \mathrm{~cm} \\
& \frac{3 \lambda}{4}-\frac{\lambda}{4}=31.25-9.75 \\
& \frac{\lambda}{2}=21.5 \\
& \lambda=43 \mathrm{~cm}^2 \\
& v=\lambda \times f \\
& v=34400 \times 10^{-2} \\
& \therefore \quad v=344 \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

For closed organ pipe, third harmonic
\(
n=\frac{(2 N-1) V}{4 \ell}=\frac{3 V}{4 \ell}(\because \mathrm{N}=2)
\)
For open organ pipe, fundamental frequenty
\(
n=\frac{N V}{2 \ell}=\frac{V}{2 \ell^{\prime}}(\because \mathrm{N}=1)
\)
According to question, \(\frac{3 V}{4 \ell}=\frac{V}{2 \ell^{\prime}}\)
\(
\Rightarrow \ell^{\prime}=\frac{4 \ell}{3 \times 2}=\frac{2 \ell}{3}=\frac{2 \times 20}{3}=13.33 \mathrm{~cm}
\)

Hint

For first resonance, \(l_1=\frac{\lambda}{4}\)
For second resonance, \(l_2=\frac{3 \lambda}{4}\)
\(\therefore \quad\left(l_2-l_1\right)=\frac{3 \lambda}{4}-\frac{\lambda}{4}\)
or \(\quad \lambda=2\left(l_2-l_1\right) \dots(i)\)
As velocity of the sound wave is given as,
\(
v=f \lambda
\)
where, \(f\) is the frequency.
\(\Rightarrow v=\left[2\left(l_2-l_1\right)\right][\) from Eq. (i)]
Here, \(f=320 \mathrm{~Hz}, l_2=0.73 \mathrm{~m}, l_1=0.20 \mathrm{~m}\)
\(\Rightarrow \quad v=2[320(0.73-0.20)]\)
\(=2 \times 320 \times 0.53\)
\(=339.2 \mathrm{~ms}^{-1} \simeq 339 \mathrm{~ms}^{-1}\)

Hint

Thinking Process Frequency of nth harmonic in a closed-end tube \(\Rightarrow \quad f=\frac{(2 n-1) \mathrm{v}}{41} \quad n=1,2,3, \ldots\)
Also, only odd harmonics exist in a closed-end tube.
Now, given two nearest harmonics are of frequency \(220 \mathrm{~Hz}\) and \(260 \mathrm{~Hz}\).
\(\therefore \quad \frac{(2 n-1) \mathrm{v}}{41}=220 \mathrm{~Hz}\dots(i)\)
The next harmonic occurs at
\(
\frac{(2 n+1) \mathrm{v}}{4}=260 \mathrm{~Hz} \dots(ii)
\)
On subtracting Eq. (i) from Eq. (ii), we get
\(
\begin{gathered}
\frac{\{(2 n+1)-(2 n-1)\} \mathrm{v}}{4l}=260-220 \\
2\left(\frac{\mathrm{v}}{4l}\right)=40 \Rightarrow \frac{\mathrm{v}}{4l}=20 \mathrm{~Hz}
\end{gathered}
\)

Hint

As we know from doppler’s effect, When both source and observer are moving towards each other, apparent frequency is given by
\(
f_a=f_0\left(\frac{v+v_0}{v-v_s}\right)
\)
where, \(f_0=\) original frequency of source
\(
\begin{aligned}
& v_s=\text { speed of source } \\
& v_0=\text { speed of observer } \\
& v=\text { speed of sound }
\end{aligned}
\)
Frequency of the horn,
\(
f_0=400 \mathrm{~Hz}
\)
Speed of the observer in the second car,
\(
v_0=16.5 \mathrm{~m} / \mathrm{s}
\)

Speed of source,
\(\begin{aligned} v_{\mathrm{s}} & =\text { speed of first car } \\ & =22 \mathrm{~m} / \mathrm{s}\end{aligned}\)
Frequency heard by the driver in the second car
\(
\begin{aligned}
f_a & =f_0\left(\frac{v+v_0}{v-v_s}\right)=400\left(\frac{340+16.5}{340-22}\right) \\
& =448 \mathrm{~Hz}
\end{aligned}
\)

Hint

According to question, we have Wavelength of transphase pulse
\(
\lambda=\frac{v}{f} \dots(i)
\)
\((v=\) velocity of the wave; \(f=\) frequency of the wave)
As we know \(v=\sqrt{\frac{T}{\mu}}\dots(ii)\)
( \(T=\) tension in the spring \(; \mu=\) mass per unit length of the rope)
From Eqs. (i) and (ii), we get
\(
\begin{aligned}
\lambda & =\frac{1}{f} \sqrt{\frac{T}{\mu}} \\
\Rightarrow \quad \lambda & \propto \sqrt{T}
\end{aligned}
\)
So, for two different case, we get
\(
\begin{aligned}
\frac{\lambda_2}{\lambda_1} & =\sqrt{\frac{T_2}{T_1}} \\
& =\sqrt{\frac{m_1+m_2}{m_2}}
\end{aligned}
\)

Hint

For an open organ pipe, \(f_n=\frac{n}{2 L} v\), where \(n=1,2,3, \ldots\)
For second overtons \(n=3, f_{2_0}=\frac{3}{2 L_1} v \dots(i)\)
\(L_1=\) length of open organ pipe For closed organ pipe \(f_n=\left(\frac{2 n+1}{4 L}\right) v\) where, \(n=0,1,2,3, \ldots\)
1st overtone for closed organ pipe, \(n=1\)
\(
f_{1_c}=\frac{3}{4 L} v \dots(ii)
\)
\(
\begin{array}{ll}
\because & f_{2_0}=f_{1_c} \Rightarrow \frac{3 \mathrm{~V}}{2 L_1}=\frac{3}{4 L} v \\
\Rightarrow & L_1=2 L
\end{array}
\)

Hint

The formula for the beat frequency is \(\mathrm{f}_{\mathrm{b}}=\mathrm{f}_2-\mathrm{f}_1\)
Where \(f_1\) and \(f_2\) are the frequency of two waves.
Now
\(
\mathrm{f}_{\mathrm{b}}=\mathrm{f}_2-\mathrm{f}_1
\)
\(
\mathrm{f}_{\mathrm{b}}=(\mathrm{n}+1)-(\mathrm{n}-1)
\)
\(
\mathrm{f}_{\mathrm{b}}=(\mathrm{n}+1-\mathrm{n}+1)
\)
\(
\mathrm{f}_{\mathrm{b}}=2
\)
Thus the number of beats produced per second will be 2.

Hint

According to Doppler’s effect in sound, Frequency of sound that the observer hear in the echo reflected from the cliff is given by
\(
f^{\prime}=\left(\frac{v}{v-v_s}\right) f
\)
where \(f=\) original frequency of source:
\(v=\) velocity of sound
\(v_s=\) velocity of source
So, \(f^{\prime}=\left(\frac{330}{330-15}\right) 800=838 \mathrm{~Hz}\)

Hint

For a closed organ pipe first minimum resonating length
\(
\begin{aligned}
& \mathrm{L}_1=\frac{\lambda}{4}=50 \mathrm{~cm} \\
& \therefore \quad \text { Next or second resonating length, } \mathrm{L}_2 \\
& \Rightarrow \frac{3 \lambda}{4}=150 \mathrm{~cm}
\end{aligned}
\)

If an open pipe is half submerged in water, it will become a closed organ pipe of length half that of an open pipe. Its fundamental frequency will become
\(f^{\prime}=\frac{V}{4\left(\frac{\ell}{2}\right)}=\frac{V}{2 \ell}\) equal to that of open pipe.

Hint

The fundamental frequency of closed organ pipe,
\(
f_{\mathrm{c}}=\frac{\mathrm{v}}{4 l_c}
\)
The fundamental frequency of open organ pipe
\(
\mathrm{f}_o=\frac{\mathrm{v}}{2 l_0}
\)
Second overtone frequency of open organ pipe
\(
=\frac{3 \mathrm{v}}{2 l_0}
\)
From question, \(\frac{\mathrm{v}}{4 l_c}=\frac{3 \mathrm{v}}{2 l_0}\)
\(
\Rightarrow l_0=6 l_c=6 \times 20=120 \mathrm{~cm}
\)

Hint

Resonant frequencies, for a string fixed at both ends, will be \(\mathrm{f}_{\mathrm{n}}=\frac{\mathrm{n} v}{2 \mathrm{~L}}\) where, \(n=1,2,3 \ldots\).
The difference between two consecutive resonant frequency is,
\(
\mathrm{f}_{\mathrm{n}+1}-\mathrm{f}_{\mathrm{n}} \Rightarrow \frac{(\mathrm{n}+1) v}{2 \mathrm{~L}}-\frac{\mathrm{nv}}{2 \mathrm{~L}}=\frac{v}{2 L}
\)
which is also the lowest resonant frequency \((n=1)\).
Thus, for the given string the lowest resonant frequency will be
\(
\begin{aligned}
& =420 \mathrm{~Hz}-315 \mathrm{~Hz} \\
& =105 \mathrm{~Hz}
\end{aligned}
\)

Alternate:
The difference between any two successive frequencies will be ‘ \(n\) ‘
According to question, \(\mathrm{n}=420-315=105 \mathrm{~Hz}\) So the lowest frequency of the string is \(105 \mathrm{~Hz}\)

Hint

Here, original frequency of sound, \(f_0=100[latex] [latex]\mathrm{Hz}\)
Speed of source \(\mathrm{V}_{\mathrm{s}}=19.4 \cos 60^{\circ}=9.7\)
From Doppler’s formula
\(
\begin{aligned}
& \mathrm{f}^{\prime}=\mathrm{f}_0\left(\frac{\mathrm{V}-\mathrm{V}_0}{\mathrm{~V}-\mathrm{V}_{\mathrm{s}}}\right) \\
& \mathrm{f}^{\prime}=100\left(\frac{\mathrm{V}-0}{\mathrm{~V}-(+9.7)}\right) \\
& \mathrm{f}^{\prime}=100 \frac{\mathrm{V}}{\mathrm{V}\left(1-\frac{9.7}{\mathrm{~V}}\right)}=\frac{100}{\left(1-\frac{9.7}{330}\right)} \\
& =103 \mathrm{~Hz} \\
& \text { Apparent frequency } \mathrm{f_a}=103 \mathrm{~Hz}
\end{aligned}
\)

Hint

In this problem, the fundamental frequencies of each part could be found. The fundamental frequency of the complete wire could be found. One should check each option for the given values.
\(
n=\frac{1}{2 l} \sqrt{\frac{T}{m}}
\)
or, \(n \propto \frac{1}{l}\) or \({n} l=\) constant, \(K\)
\(
\begin{aligned}
& \therefore n_1 l_1=K, \\
& n_2 l_2=K, n_3 l_3=K
\end{aligned}
\)
Also, \(l=l_1+l_2+l_3\)
or, \(\frac{K}{n}=\frac{K}{n_1}+\frac{K}{n_2}+\frac{K}{n_3}\)
or, \(\frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}\)

Hint

In case of closed organ pipe frequency,
\(
\begin{aligned}
\mathrm{f}_{\mathrm{n}}= & (2 \mathrm{n}+1) \frac{v}{4 l} \\
\text { for } & \mathrm{n}=0, \mathrm{f}_0=100 \mathrm{~Hz} \\
\mathrm{n} & =1, \mathrm{f}_1=300 \mathrm{~Hz} \\
\mathrm{n} & =2, \mathrm{f}_2=500 \mathrm{~Hz} \\
\mathrm{n} & =3, \mathrm{f}_3=700 \mathrm{~Hz} \\
\mathrm{n} & =4, \mathrm{f}_4=900 \mathrm{~Hz} \\
\mathrm{n} & =5, \mathrm{f}_5=1100 \mathrm{~Hz} \\
\mathrm{n} & =6, \mathrm{f}_6=1300 \mathrm{~Hz}
\end{aligned}
\)
Hence possible natural oscillation whose frequencies \( < 1250 \mathrm{~Hz}=6(\mathrm{n}=0,1,2,3,4,5)\)

Hint

According to Doppler’s effect Apparent frequency,
\(\Rightarrow\) As both observer and source are moving, we can use the formula of apparent frequency as
\(
\begin{aligned}
& f=f_0\left(\frac{v+v_0}{v+v_s}\right) \\
& =1392\left[\frac{343+10}{343+5}\right] \\
& {\left[\because v_0=36 \mathrm{~km} / \mathrm{h}=10 \mathrm{~m} / \mathrm{s}\right. \text { and }} \\
& =1392\left[\frac{353}{348}\right]=1412 \mathrm{~Hz}
\end{aligned}
\)

Hint

The standard equation of a wave travelling along \(x\)-axis (+ve direction) is given by, \(\mathrm{Y}=\mathrm{A} \sin (k x-\omega t)\) where, angular frequency, \(\omega=2 \pi f\)
\(
\begin{array}{lc}
\Rightarrow \quad \frac{2 \pi}{\pi}=2 \quad\left[\because f=\frac{1}{\pi}\right] \\
\text { angular wave number, } k=\frac{2 \pi}{\lambda} \Rightarrow & \frac{2 \pi}{2 \pi}=1 \\
& {[\because \lambda=2 \pi]}
\end{array}
\)
\(
\therefore \mathrm{Y}=1 \sin (x-2 t)[\because \text { Amplitude, } \mathrm{A}=1 \mathrm{~m}]
\)

If the sign between \(t\) and \(x\) terms is negative the wave is propagating along \(+(\) ve) \(X\)-axis and if the sign is positive then wave moves in \(-(\mathrm{ve}) X\)-axis direction.

Hint

Pressure change will be minimum at both ends. In fact, pressure variation is maximum at \(\ell / 2\) because the displacement node is pressure antinode.

Hint

When sounded with a source of known frequency fundamental frequency \(=250 \pm 4 \mathrm{~Hz}=254 \mathrm{~Hz}\) or \(246 \mathrm{~Hz}\) \(2^{\text {nd }}\) harmonic if unknown frequency (suppose) \(254 \mathrm{~Hz}=2 \times 254=508 \mathrm{~Hz}\) As it gives 5 beats
\(
\therefore 508+5=513 \mathrm{~Hz}
\)
Hence, unknown frequency is \(254 \mathrm{~Hz}\)

Hint

From formula, \(f=\frac{1}{x} \sqrt{\frac{T}{m}}\)
\(
\begin{aligned}
& \Rightarrow f \propto \frac{1}{l} \\
& \therefore l_1: l_2: l_3=\frac{1}{f_1}: \frac{1}{f_2}: \frac{1}{f_3} \\
& =f_2 f_3: f_1 f_3: f_1 f_2 \quad\left[\text { Given: } f_1: f_2: f_3=1: 3: 5\right] \\
& =15: 5: 3
\end{aligned}
\)
Therefore the positions of the two bridges below the wire are
\(\frac{15 \times 100}{15+5+3} \mathrm{~cm}\) and \(\frac{15 \times 100+5 \times 100}{15+5+3} \mathrm{~cm}\)
i.e., \(\frac{1500}{23} \mathrm{~cm}, \frac{2000}{23} \mathrm{~cm}\)

Hint

\(
\begin{aligned}
& \text { (b) } \frac{\Delta f}{f}=\frac{v}{c} \\
& \Rightarrow \frac{(\text { Beats) } / 2}{f}=\frac{v}{c} \\
& \Rightarrow \text { Beats }=\frac{2 f v}{c}=4
\end{aligned}
\)

Hint

\(2 \pi f_1=600 \pi\)
\(
\begin{aligned}
& f_1=300 \\
& 2 \pi f_2=608 \pi \\
& f_2=304 \\
& \left|f_1-f_2\right|=4 \text { beats } \\
& \frac{l_{\max }}{l_{\min }}=\frac{\left(A_1+A_2\right)^2}{\left(A_1+A_2\right)^2}=\frac{(5+4)^2}{(5-4)^2}=\frac{81}{1}
\end{aligned}
\)
where \(A_1, A_2\) are amplitudes of given two sound wave.

Hint

Frequency of the echo detected by the driver of the train is
(According to Doppler effect in sound)
\(
f^{\prime}=\left(\frac{v+u}{v-u}\right) f
\)
where \(f=\) original frequency of source of sound \(f^{\prime}=\) Apparent frequency of source because of the relative motion between source and observer.
\(
f^{\prime}=\left(\frac{330+220}{330-220}\right) 1000=5000 \mathrm{~Hz}
\)
The velocity of sound direction is always taken from source to observer. All the velocities in the direction of \(v\) are positive and in the opposite direction of \(v\) are taken negative.

Hint

According to question,
\(
y_1=a \sin (\omega t+k x+0.57)
\)
and \(y_2=a \cos (\omega t+k x)\)
or \(\quad y_2=a \sin \left(\frac{\pi}{2}+\omega t+k x\right)\)
As phase difference,
\(
\begin{aligned}
\Delta \phi & =\phi_2-\phi_1=\frac{\pi}{2}-0.57 \\
& =1.57-0.57=1 \mathrm{rad}
\end{aligned}
\)

Hint

Velocity of a wave is given by
\(
v=n \lambda
\)
\(
\left[\begin{array}{l}
n=\text { frequency of wave } \\
\lambda=\text { wavelength of wave }
\end{array}\right]
\)
So, for two different cases,
\(
\begin{aligned}
v_1 & =n_1 \lambda_1 \\
v_2 & =n_2 \lambda_2 \\
\lambda_2 & =\lambda_1 \frac{v_2}{v_1} \\
& =\lambda_1 \times \frac{3500}{350}=\lambda_1 \times 10 \\
\lambda_2 & =10 \lambda_1 \quad\left[\because n_1=n_2\right]
\end{aligned}
\)

Hint

For fundamental mode, \(\mathrm{f}=\frac{1}{2 \ell} \sqrt{\frac{T}{\mathrm{~m}}}\) Taking logarithm on both sides, we get
\(
\begin{aligned}
& \Rightarrow \log f=\log \left(\frac{1}{2 \ell}\right)+\log \left(\sqrt{\frac{T}{\mu}}\right) \\
& \Rightarrow \log \left(\frac{1}{2 \ell}\right)+\frac{1}{2} \log \left(\frac{T}{\mu}\right)
\end{aligned}
\)
or \(\log f=\log \left(\frac{1}{2 \ell}\right)+\frac{1}{2}[\log T-\log \mu]\)
Differentiating both sides, we get \(\frac{\mathrm{df}}{\mathrm{f}}=\frac{1}{2} \frac{\mathrm{dT}}{\mathrm{T}}\) (as \(\ell\) and \(\mu\) are constants)
\(
\Rightarrow \frac{\mathrm{dT}}{\mathrm{T}}=2 \times \frac{\mathrm{df}}{\mathrm{f}}
\)
Here \(\mathrm{df}=6\)
\(
\begin{aligned}
& \mathrm{f}=600 \mathrm{~Hz} \\
& \therefore \frac{\mathrm{dT}}{\mathrm{T}}=\frac{2 \times 6}{600}=0.02
\end{aligned}
\)

Hint

For a particle executing SHM
Acceleration \(a \propto-\omega^2\) displacement (x) \(\quad \ldots(i)\)
\(
\text { Given } \quad x=a \sin ^2 \omega t \dots(ii)
\)
Differentiating the above equation w.r.t, we get \(\frac{d x}{d t}=2 a \omega(\sin \omega t)(\cos \omega t)\)
Again differentiating, we get
\(
\begin{aligned}
\frac{d^2 x}{d t^2} & =a=2 a \omega^2\left[\cos ^2 \omega t-\sin ^2 \omega t\right] \\
& =2 a \omega^2 \cos 2 \omega t
\end{aligned}
\)
The given equation does not satisfy the condition for SHM [Eq. (i)]. Therefore, motion is not simple harmonic.

Hint

\(y=A \sin (\omega t-k x)\)
\(
v_w=\frac{\lambda}{T}=\frac{\omega \lambda}{2 \pi}
\)
Particle velocity,
\(
\begin{aligned}
& v_p=\frac{d y}{d t}=A \omega \cos (\omega t-k x) \\
& \therefore \quad v_{p \max }=A \omega \\
& \text { wave velocity }=\frac{\omega}{k} \\
& \therefore \quad A \omega=\frac{\omega}{k} \\
& \text { i.e., } A=\frac{1}{k} \text { But } k=\frac{2 \pi}{\lambda} \\
& \therefore \quad \lambda=2 \pi A
\end{aligned}
\)

Hint

Frequency of tuning fork \(=512 \mathrm{~Hz}\)
Number ob beats \(=4\)
reduced beats \(=2\)
\(\because f \propto \sqrt{T}\)
beats \(=n_f-n_p\)
\(4=512-n_p \Rightarrow n_p=508 \mathrm{~Hz}\)

Hint

Given, amplitude of wave, \(A=2 \mathrm{~cm}\) direction \(=+\) ve \(x\) direction
Velocity of wave
\(
v=128 \mathrm{~ms}^{-1}
\)
and length of string, \(5 \lambda=4\)
We know that,
\(
k=\frac{2 \pi}{\lambda}=\frac{2 \pi \times 5}{4}=7.85
\)
\(
\begin{aligned}
& \text { and } \quad v=\frac{\omega}{k}=128 \mathrm{~ms}^{-1} \\
& \text { [ } \omega=\text { Angular frequency }] \\
& \Rightarrow \omega=v \times k=128 \times 7.85=1005 \\
&
\end{aligned}
\)
As, the wave travelling towards \(+x\)-axis is given by
\(
\begin{aligned}
y & =A \sin (k x-\omega t) \\
\text { So, } y & =2 \sin (7.85 x-1005 t) \\
y & =(0.02) m \sin (7.85 x-1005 t)
\end{aligned}
\)

Hint

From the given condition
\(\mathrm{f}^{\prime}\) is the apparent frequency received by an observer at the hill. \(f^{\prime \prime}\) is the frequency of the reflected sound as heard by driver.
\(
\begin{aligned}
f^{\prime} & =\frac{v}{v-30} f \text { and } \\
f^{\prime \prime} & =\frac{v+30}{v} f^{\prime} \Rightarrow \frac{v+30}{v-30} f=\frac{360}{300} \times 600 \\
& =720 \mathrm{~Hz}
\end{aligned}
\)

Hint

The frequency of vibration of a string is given by, \(f=\frac{1}{2l} \sqrt{\frac{T}{m}}\) where \(m\) is mass per unit length.
\(
\begin{aligned}
& \mathrm{f}_1=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}, \mathrm{f}_2=\frac{1}{2 {l}_2} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\
& \mathrm{f}_2-\mathrm{f}_1=\frac{1}{2} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \frac{\left({l}_1-{l}_2\right)}{{l}_1 {l}_2}
\end{aligned}
\)
\(
\begin{aligned}
& \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\sqrt{\frac{20}{10^{-3}}}=\sqrt{2} \times 10^2=1.414 \times 100 \\
& =141.4
\end{aligned}
\)
\(
\begin{aligned}
& \frac{l_1-l_2}{l_1 l_2}=\frac{(51.6-49.1) \times 10^2}{51.6 \times 49.1} \\
& =\frac{2.5 \times 10^2}{50 \times 50}=\frac{1}{10} \\
& \therefore \quad \mathrm{f}_2-\mathrm{f}_1=\frac{1}{2} \times 141.4 \times \frac{1}{10}=7 \text { beats }
\end{aligned}
\)

Hint

\(y=0.25 \sin (10 \pi x-2 \pi t)\)
Comparing this equation with the standard wave equation \(y=a \sin (k x-\omega t)\)
We get, \(k=10 \pi\)
\(
\Rightarrow \frac{2 \pi}{\lambda}=10 \pi \Rightarrow \lambda=0.2 \mathrm{~m}
\)
And \(\omega=2 \pi\) or, \(2 \pi f=2 \pi \Rightarrow f=1 \mathrm{~Hz}\). The sign inside the bracket is negative, hence the wave travels in +ve \(x\)-direction.

Hint

As intensity is directly proportional to the square of amplitude
i.e. \(\quad I \propto a^2\)
So, maximum intensity is given by
\(
I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2
\)
\(\left[\begin{array}{l}I_1, I_2 \text { are intensities } \\ \text { of two waves }\end{array}\right]\)
and \(I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)
\(
\begin{aligned}
\therefore I_{\max }+I_{\min } & =\left(\sqrt{I_1}+\sqrt{I_2}\right)^2+\left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \\
& =2\left(I_1+I_2\right)
\end{aligned}
\)

Hint

Sound waves in air are longitudinal and the light waves are transverse.

Note: Sound waves are longitudinal waves and require a medium for their propagation.
Light waves i.e., electromagnetic waves are transverse in nature and do not require a medium for their propagation.

In a longitudinal wave, the particles of the medium oscillate about their mean or equilibrium position along the direction of propagation of the wave itself. Sound waves are longitudinal in nature. In transverse wave, the particles of the medium oscillate about their mean or equilibrium position at right angles to the direction of propagation of wave itself. Light waves being electromagnetic are transverse waves.

Hint

The time gap between the initial direct note and the reflected note up to the minimum audibility level is called reverberation time. Sabine has shown that standard reverberation time for an auditorium is given by the formula \(T_R=K \frac{V}{\alpha S}\)
Where, \(V\) is the volume of the auditorium, \(\mathrm{S}\) is the
surface area and \(\alpha\) is the absorption coefficients. So, \(T_R=\frac{K \cdot V}{\alpha S}=1\) (given)
So, for two different cases of reverberation.
\(
\begin{aligned}
\therefore \quad & \frac{T_R^{\prime}}{T_R}=\frac{V^{\prime}}{S^{\prime}} \times \frac{S}{V} \\
= & \frac{(2)^3}{(2)^2}=\frac{8}{4}=2
\end{aligned}
\)
\(
\text { Hence, } \quad T_R^{\prime}=2 T_R=2 \times 1=2 \mathrm{~s}
\)

Hint

Speed of a wave represented by the equation
\(y(x, t)=\mathrm{A} \sin (k x-\omega t+\phi)\) is \(v=\frac{\omega}{k}\)
By comparison, \(\omega=4 \pi ; k=0.5 \pi\)
\(
\text { Speed of transverse wave, }
\)
\(
v=\frac{\omega}{k}=\frac{4 \pi}{0.5 \pi}=8 \mathrm{~m} / \mathrm{sec}
\)

Hint

Let \(\lambda_1=5.0 \mathrm{~m}, \mathrm{v}=330 \mathrm{~m} /\) sand \(\lambda_2=5.5 \mathrm{~m}\) The relation between frequency \((f)\), wavelength \((\lambda)\) and velocity \((v)\) is given by
\(
\begin{aligned}
\Rightarrow & v=f \lambda \\
& f=\frac{v}{\lambda}
\end{aligned}
\)
The frequency corresponding to wavelength \(\lambda_1\),
\(
f_1=\frac{v}{\lambda_1}=\frac{330}{5.0}=66 \mathrm{~Hz}
\)
The frequency corresponding to wavelength \(\lambda_2\),
\(
f_2=\frac{v}{\lambda_2}=\frac{330}{5.5}=60 \mathrm{~Hz}
\)
Hence, the number of beats per second
\(
=f_1-f_2=66-60=6
\)

Hint

Intensity of sound, \(I=\frac{p}{4 \pi r^2}\)
\(
\left[\begin{array}{c}
p=\text { pressure of sound waves } \\
r=\text { distance between source } \\
\text { and the point }
\end{array}\right]
\)
or \(\quad I \propto \frac{1}{r^2}\)
or \(\quad \frac{I_1}{I_2}=\left(\frac{r_2}{r_1}\right)^2\)
Here, \(\quad r_1=2 m, r_2=3 \mathrm{~m}\)
Substituting the values, we have
\(
\frac{I_1}{I_2}=\left(\frac{3}{2}\right)^2=\frac{9}{4}
\)

Hint

Equation of progressive wave is given by \(y=A \sin 2 \pi f t\)
Given \(y_1=4 \sin 500 \pi t\) and \(y_2=2 \sin 506 \pi t\).
Comparing the given equations with the equation of progressive wave, we get
\(2 f_1=500 \Rightarrow f_1=250\)
\(2 f_2=506 \Rightarrow f_2=253\)
Beats \(=f_2-f_1=253-250=3\) beats \(/ \mathrm{sec}\) \(=3 \times 60=180\) beats \(/\) minute.

Hint

The given waves are
\(
y_1=10^{-8} \sin \left[100 t+\left(\frac{x}{50}\right)+0.5\right] m \ldots(i)
\)
\(
\text { and } y_2=10^{-8} \cos \left[100 t+\left(\frac{x}{50}\right)\right] \mathrm{m} \dots(ii)
\)
Eq. (ii)can be written as
\(
\begin{aligned}
\Rightarrow y_2=10^{-8} \sin & {\left[100 t+\left(\frac{x}{50}\right)+\frac{\pi}{2}\right] \mathrm{m} } \\
& {\left[\because \sin \left(\frac{\pi}{2}+\theta\right)=\cos \theta\right] }
\end{aligned}
\)
Hence, the phase difference between the waves is
\(
\begin{aligned}
\Delta \phi & =\left(\frac{\pi}{2}-0.5\right) \mathrm{rad} \\
& =\left(\frac{3.14}{2}-0.5\right) \mathrm{rad} \\
& =(1.57-0.5) \mathrm{rad} \\
& =(1.07) \mathrm{rad}
\end{aligned}
\)

Hint

Let \(f^{\prime}\) be the frequency of sound heard by cliff.
\(
\therefore f=\frac{v f}{v-v_c} \dots(1)
\)
Now, for the reflected wave, cliff acts as a source,
\(
\begin{aligned}
& \therefore 2 f=\frac{f^{\prime}\left(\mathrm{v}+\mathrm{v}_{\mathrm{c}}\right)}{v} \\
& 2 f=\frac{\left(v+v_{\mathrm{c}}\right) f}{v-v_c} \dots(2)
\end{aligned}
\)
\(
\Rightarrow 2 v-2 v_c=v+v_{\mathrm{c}} \text { or } \frac{v}{3}=v_c
\)

Hint

When an observer moves towards a stationary source of sound, then apparent frequency heard by the observer increases. The apparent frequency heard in this situation
\(
f^{\prime}=\left(\frac{v+v_a}{v-v_s}\right) f
\)
As source is stationary hence, \(v_s=0\)
\(
\therefore \quad f^{\prime}=\left(\frac{v+v_a}{v}\right) f
\)
Given, \(\quad v_a=\frac{v}{5}\)
Substituting in the relation for \(f^{\prime}\), we have
\(
f^{\prime}=\left(\frac{v+v / 5}{v}\right) f=\frac{6}{5} f=1.2 f
\)

The motion of the observer does not affect the wavelength reaching the observer, hence, the wavelength remains \(\lambda\).

Hint

The general expression of travelling wave can be written as
\(
y=a \sin (\omega t \pm k x) \dots(i)
\)
For travelling wave along positive \(x\)-axis we should use minus (-) sign only.
\(\therefore \quad y=a \sin (\omega t-k x)\)
but \(\omega=\frac{2 \pi v}{\lambda}\) and \(k=\frac{2 \pi}{\lambda}\)
So, \(\quad y=a \sin \frac{2 \pi}{\lambda}(v t-x) \dots(ii)\)
Given, \(a=02 \mathrm{~m}, v=360 \mathrm{~m} / \mathrm{s}, \lambda=60 \mathrm{~m}\), Substituting in Eq. (ii), we have
\(
y=0.2 \sin \frac{2 \pi}{60}(360 t-x)
\)
or \(\quad y=0.2 \sin 2 \pi\left(6 t-\frac{x}{60}\right)\)

Hint

Velocity of source, \(v_s=r \omega\)
\(
=0.50 \times 20=10 \mathrm{~ms}^{-1}
\)
The frequency of sound observed by the observer will be minimum when he is at point \(A\). Thus, at this point minimum frequency of source as observed by observer is
\(
\begin{aligned}
f_{\min } & =\left(\frac{v}{v+v_s}\right) f \\
f_{\min } & =\frac{340}{340+10} \times 385 \\
& =\frac{34}{35} \times 385=34 \times 11=374 \mathrm{~Hz}
\end{aligned}
\)

Hint

The given wave equation is
\(
y=a \sin \left(100 t-\frac{x}{10}\right)
\)
Compare it with the standard wave equation, we obtain
\(
\omega=100, k=\frac{1}{10}
\)
Velocity of the wave,
\(
\begin{aligned}
v & =\frac{\omega}{k}=\frac{100}{1 / 10} \\
& =100 \times 10=1000 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

When a wave enters from one medium to another, its frequency remains unchanged, i.e.
\(n_1=n_2\) but wavelength, intensity and velocity get changed.

Hint

The given progressive waves are
\(
\begin{aligned}
& y_1=a \sin \left(\omega t+\phi_1\right) \\
& y_2=a \sin \left(\omega t+\phi_2\right)
\end{aligned}
\)
The resultant of two waves is
\(
\begin{aligned}
y & =y_1+y_2 \\
& =a\left[\sin \left(\omega t+\phi_1\right)+\sin \left(\omega t+\phi_2\right)\right]
\end{aligned}
\)
If \(A\) is the amplitude of resultant wave, then
\(
\begin{aligned}
& A=a \quad \text { (given) } \\
& \therefore \quad A^2=a^2+a^2+2 a^2 \cos \phi \\
& \text { or } a^2=a^2+a^2+2 a^2 \cos \phi \\
& \text { or } \cos \phi=-\frac{1}{2}=\cos 120^{\circ} \\
& \therefore \quad \phi=120^{\circ}=\frac{2 \pi}{3} \\
& \text { Thus, } \quad \phi_1-\phi_2=\frac{2 \pi}{3} \\
&
\end{aligned}
\)

Hint

\(
f^{\prime}=f\left(\frac{v-v_o}{v+v_s}\right)=f\left(\frac{340-11}{340+10}\right)=1950 \Rightarrow f=2068 \mathrm{~Hz}
\)

Hint

The frequency of vibrations of the string is
\(
n=\frac{1}{2} \sqrt{\frac{g}{l}} \dots(i)
\)
Given, \(n_A=2 n_B\)
\(\therefore \quad \frac{1}{2} \sqrt{\frac{g}{l_A}}=2 \cdot \frac{1}{2} \sqrt{\frac{g}{l_B}}\)
or
\(
\frac{1}{l_A}=\frac{4}{l_B} \text { or } l_B=4 l_A
\)
It is obvious from Eq. (i), the frequency of vibrations of strings does not depend on their masses.

Hint

Frequency received by listener from the rear source,
\(
n^{\prime}=\frac{v-u}{v} \times n=\frac{v-u}{v} \times \frac{v}{\lambda}=\frac{v-u}{\lambda}
\)
Frequency received by listener from the front source,
\(
n^{\prime \prime}=\frac{v+u}{v} \times \frac{v}{\lambda}=\frac{v+u}{\lambda}
\)
No. of beats \(=n^{\prime \prime}-n^{\prime}\)
\(
=\frac{v+u}{\lambda}-\frac{v-u}{\lambda}=\frac{v+u-v+u}{\lambda}=\frac{2 u}{\lambda}
\)

Hint

Beats produced due to the two frequencies is given by
\(
n_1-n_2
\)
where, \(n_1\) and \(n_2\) are the frequencies of two waves.
Here, number of beats \(=12 / \mathrm{s}\)
or
\(
\begin{aligned}
& \lambda_1=50 \mathrm{~cm}=0.50 \mathrm{~m} \\
& \lambda_2=51 \mathrm{~cm}=0.51 \mathrm{~m} \\
& \mathrm{n}_1-n_2=12 \\
& \frac{v}{\lambda_1}-\frac{v}{\lambda_2}=12 \quad\left[n=\frac{v}{\lambda}\right]
\end{aligned}
\)
or \(\quad v\left(\frac{\lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right)=12\)
or
\(
v=\frac{12 \lambda_1 \lambda_2}{\lambda_2-\lambda_1}
\)
\(\therefore \quad v=\) speed of sound
\(
\begin{aligned}
& =\frac{12 \times 0.50 \times 0.51}{(0.51-0.50)}=\frac{12 \times 0.50 \times 0.51}{0.01} \\
& =306 \mathrm{~m} / \mathrm{s} \\
& \text { Thus, speed of sound is } 306 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

Time taken to move from maximum to zero displacement \(=\frac{T}{4}\)
\(\therefore\) Time period \(T=4 \times 0.170\) second
\(\therefore\) Frequency, \(f=\frac{1}{T}=\frac{1}{4 \times 0.170}=1.47 \mathrm{~Hz}\)

Hint

\(y=y_0 \sin \frac{2 \pi}{\lambda}(v t-x)\)
Particle velocity,
\(
\frac{d y}{d t}=y_0 \times \frac{2 \pi}{\lambda} v \cos \frac{2 \pi}{\lambda}(v t-x)
\)
Maximum particle velocity \(=y_0 \times \frac{2 \pi v}{\lambda}\)
Wave velocity \(=v \quad\) [given]

Maximum particle velocity \(=2 \times\) wave velocity
So, \(y_0 \times \frac{2 \pi v}{\lambda}=2 v\)
\(\lambda=\pi \cdot y_0\)

Hint

The given standing wave is shown in the figure.

As the length of one loop or segment is \(\frac{\lambda}{2}\), so the length of 2 segments is \(2\left(\frac{\lambda}{2}\right)\).
So, according to the question
\(\therefore \quad 2 \frac{\lambda}{2}=121 Å \Rightarrow \lambda=1.21 Å\)

Alternate:
Let \(\ell\) be length of string
\(
\ell=\left(\frac{\lambda}{2}\right) 2 \Rightarrow \lambda=\ell
\)
Hence, the wavelength of the standing wave
\(
\Rightarrow \lambda=\ell=1.21 Å
\)

Hint

When the velocity of the source (vehicle) is perpendicular to the line joining the observer and source, then there is no Doppler effect of sound as the component of velocity either towards or away from the observer is zero. So, there is no change in apparent frequency. Therefore, \(n_1=0\).

Note:

As the source is not moving towards or away from the observer in a straight line, the Doppler’s effect will not be observed by the observer.

If the source and observer both are relatively at rest and if the speed of sound is increased then the frequency heard by the observer will not change.

Hint

The fundamental frequency of open pipe,
\(
f=\frac{v}{2 l}
\)
When half of tube is filled with water, then the length of air column becomes half \(\left(l^{\prime}=\frac{l}{2}\right)\) and the pipe becomes closed.
So, new fundamental frequency
\(
f^{\prime}=\frac{v}{4 l^{\prime}}=\frac{v}{4\left(\frac{l}{2}\right)}=\frac{v}{2 l}
\)
Clearly \(f^{\prime}=f\).

Hint

A pulse of a wave train when travels along a stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium and the reflected ray suffers a phase change of \(\pi\) with the incident wave and wave velocity after reflection will reverse.

Hint

In the case of standing wave, the length of one segment is \(\frac{\lambda}{2}\). There are 5 segments and total length of string is \(10 \mathrm{~m}\). \(\therefore \quad 5 \frac{\lambda}{2}=10 \Rightarrow \lambda=4 \mathrm{~m}\) Frequency, \(n=\frac{v}{\lambda}=\frac{20}{4}=5 \mathrm{~Hz}\)
\(
(\because v=20 \mathrm{~m} / \mathrm{s})
\)
Standing wave is an example of interference. Destructive interference means node and constructive interference means antinode.

Hint

Given speed of wave \((v)=960 \mathrm{~m} / \mathrm{s}\) Frequency of wave \((f)=3600 / \mathrm{min}\) \(=\frac{3600}{60} \mathrm{rev} / \mathrm{sec}=60 \mathrm{rev}\) per sec.
Wavelength of waves \((\lambda)=\frac{v}{f}=\frac{960}{60}=16 \mathrm{~m}\).

Hint

\(
y=60 \cos (180 t-6 x) \dots(1)
\)
\(
\omega=180, k=6 \Rightarrow \frac{2 \pi}{\lambda}=6
\)
Wave velocity,
\(
v=\frac{\omega}{k}=\frac{2 \pi}{T} \times \frac{\lambda}{2 \pi}=\frac{180}{6}=30 \mathrm{~m} / \mathrm{s}
\)
Differentiating (1) w.r.t. \(t\),
Particle velocity,
\(
v_p=\frac{d y}{d t}=-60 \times 180 \sin (180 t-6 x)
\)
\(
\begin{aligned}
& v_{p \max }=60 \times 180 \mu \mathrm{m} / \mathrm{s} \\
& =10800 \mu \mathrm{m} / \mathrm{s}=0.0108 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(
\frac{v_{\mathrm{p} \mathrm{max}}}{v}=\frac{0.0108}{30}=3.6 \times 10^{-4}
\)

Wave velocity \((v)\) is the distance travelled by the disturbance in one second. It only depends on the properties of the medium and is independent of time and position.
\(
V=n \lambda=f \lambda=\frac{\lambda}{T}=\frac{\omega \lambda}{2 \pi}=\frac{\omega}{k}
\)

Hint

We know that the length of pipe closed at one end for first overtone \(\left(l_1\right)=\frac{3 \lambda}{4}\) and length of the open pipe for third overtone \(\left(l_2\right)=\frac{4 \lambda}{2}=2 \lambda\)
Therefore, the ratio of lengths \(\frac{l_1}{l_2}=\frac{3 \lambda / 4}{2 \lambda}=\frac{3}{8}\) or \(l_1: l_2=3: 8\).

Note:

In an open pipe, all harmonics are present whereas in a closed organ pipe, only alternate harmonies of frequencies are present. Hence musical sound produced by an open organ pipe is sweeter than that produced by a closed organ pipe.

Hint

\(y=0.0015 \sin (62.4 x+316 t)\)
On comparing with \(y=A \sin (\omega t+k x)\)
\(
\begin{aligned}
& \omega=316, k=62.4 \\
& \Rightarrow k=\frac{2 \pi}{\lambda}=62.4 \Rightarrow \lambda=0.1 \text { unit }
\end{aligned}
\)

Hint

Velocity of sound \(=\sqrt{\frac{\gamma R T}{M}}\)
When water vapour are present in the air, the average molecular weight of air decreases, and hence velocity increases.

Hint

Since the waves are in opposite phases, the resultant intensity will be zero. 

Note: Interference phenomenon is common to sound and light. In sound, the interference is said to be constructive at points where the resultant intensity is maximum (are in phase)and destructive at points where the resultant intensity is minimum or zero (are in opposite phase).

Hint

Given: Wavelength \((\lambda)=5000 Å\) velocity of star \((v)=1.5 \times 10^6 \mathrm{~m} / \mathrm{s}\). We know that wavelength of the approaching \(\operatorname{star}\left(\lambda^{\prime}\right)=\lambda\left(\frac{c-v}{c}\right)\) or, \(\frac{\lambda^{\prime}}{\lambda}=\frac{c-v}{c}=1-\frac{v}{c}\) or, \(\frac{v}{c}=1-\frac{\lambda^{\prime}}{\lambda}=\frac{\lambda-\lambda^{\prime}}{\lambda}=\frac{\Delta \lambda}{\lambda}\) Therefore, \(\Delta \lambda=\lambda \times \frac{v}{c}=5000 \times \frac{1.5 \times 10^6}{3 \times 10^8}=25 Å\)

\(
\text { [where } \Delta \lambda=\text { Change in the wavelength] }
\)

Hint

Phase difference \(=60^{\circ}=\frac{\pi}{3}\) Path difference \(=\frac{\lambda}{2 \pi}\) (phase diff. \()\)
\(
=\frac{\lambda}{2 \pi} \times \frac{\pi}{3}=\frac{\lambda}{6}
\)

Hint

Given, speed of wave \((v)=760 \mathrm{~m} / \mathrm{s}\)
Number of waves \(=3600\)
Time, \(t=2 \mathrm{~min}=2 \times 60=120 \mathrm{~s}\)
\(\therefore\) Frequency of waves, \(f=\frac{\text { Total no. of waves }}{\text { time taken }}=\frac{3600}{120}=30 \mathrm{~Hz}\)
\(\)\therefore\(\) Wavelength of waves
\(\)
\lambda=\frac{v}{f}=\frac{760}{30}=25.3 \mathrm{~m}
[/latex

Hint

Wavelength of a wave is the length of one wave. It is equal to the distance travelled by the wave during one complete cycle, wavelength of a wave is given by
\(
\lambda=\frac{v}{f}
\)
where \(v=\) velocity of wave (sound)
\(f=\) frequency of wave (sound)
Given, \(\quad v=1.7 \times 10^3 \mathrm{~m} / \mathrm{s}\)
\(
\begin{aligned}
f  =4.2 \times 10^8 \mathrm{~Hz} \\
\therefore \quad \lambda  =\frac{1.7 \times 10^3}{4.2 \times 10^6}=4 \times 10^{-4} \mathrm{~m}
\end{aligned}
\)

Hint

Two waves are said to be coherent when they have the same frequency, amplitude, and constant phase difference.

Hint

Given, \(y=0.5 \sin \frac{2 \pi}{3.2}(64 \mathrm{t}-\mathrm{x})\)
Standard equation of the wave is :
\(
y=a \sin \frac{2 \pi}{\lambda}(v t-x)
\)
Comparing the given equation with the standard equation, we get \(v=64\) and \(\lambda=3.2\). Therefore, frequency \(=\frac{64}{3.2}=20 \mathrm{~Hz}\).

Hint

Distance between two successive
\(
\text { nodes }=\frac{\lambda}{2}
\)
but we know that, \(v=f \lambda\)
\(\therefore \quad \frac{\lambda}{2}=\frac{v}{2 f}\)
Given, \(\quad v=20 \mathrm{~m} / \mathrm{s}\)
frequency, \(\quad f=n\)
So, \(\quad \frac{\lambda}{2}=\frac{20}{2 n}=\frac{10}{n}\)

Hint

requency of first source with 5 beats \(/ \mathrm{sec}=100 \mathrm{~Hz}\) and frequency of second source with \(5 \mathrm{beats} / \mathrm{sec}=205 \mathrm{~Hz}\). The frequency of the first source \(=100 \pm 5=105\) or \(95 \mathrm{~Hz}\).
Therefore, frequency of second harmonic of source \(=210 \mathrm{~Hz}\) or \(190 \mathrm{~Hz}\).
As the second harmonic gives 5 beats/ second with the sound of frequency \(205 \mathrm{~Hz}\), therefore, frequency of the second harmonic source should be \(210 \mathrm{~Hz}\) or frequency of source \(=105 \mathrm{~Hz}\).

Hint

Velocity of sound \(\left(c_s\right)\) is given by
\(
c_s=\sqrt{\frac{\gamma p}{\rho}} \dots(i)
\)
where, \(p\) is pressure, \(\rho\) is density and \(\gamma\) is atomicity of gas or ratio of \(C_p\) and \(C_v\).
RMS velocity of gas molecules is given by
\(
c=\sqrt{\left(\frac{3 p}{\rho}\right)} \dots(ii)
\)
From Eqs. (i) and (ii)
\(
\begin{aligned}
\frac{c_s}{c} & =\sqrt{\frac{\gamma p}{\rho} \times \frac{\rho}{3 p}}=\sqrt{\frac{\gamma}{3}} \\
\Rightarrow \quad c_s & =c \times \sqrt{\left(\frac{\gamma}{3}\right)}
\end{aligned}
\)

Hint

\(y=A \sin (a t-b x+c)\) represents a wave, where a may correspond to \(\omega\) and \(b\) may correspond to \(K\).

Explanation:

For it to be a wave equation it must satisfy this
\(
\frac{\partial^2 y}{\partial t^2}=k^2 \frac{\partial^2 y}{\partial x^2}
\)

\(
\begin{aligned}
& \frac{\partial^2 y}{\partial t^2}=\frac{\partial A a \cos (a t-b x+c)}{\partial t}=-A a^2 \sin (a t-b x+c) \\
& \frac{\partial^2 y}{\partial x^2}=+A b^2 \sin (a t-b x+c)
\end{aligned}
\)

If we take these two results it comes in the format that defines a wave equation.

Hint

As fixed end is a node, therefore distance between two consecutive nodes
\(
\begin{aligned}
& =\frac{\lambda}{2}=10 \mathrm{~cm} \\
& {[\lambda=\text { wavelength of wave sent }] } \\
\Rightarrow \quad \lambda & =20 \mathrm{~cm}=0.2 \mathrm{~m}
\end{aligned}
\)
As we know, \(v=f \lambda\)
\(
\begin{aligned}
& \qquad\left[\begin{array}{l}
v=\text { velocity of wave } \\
f=\text { frequency of wave }
\end{array}\right] \\
& \therefore \quad v=100 \times 0.2=20 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

\(
y=a \sin (\omega t) \cos (k x) \dots(i)
\)
Given equation is
\(
y=a \sin (100 t) \cos (0.01 x) \dots(ii)
\)
Comparing Eqs. (i) and (ii)
\(
\omega=100 \text { and } k=0.01
\)
\(\therefore\) Velocity of wave is
\(
\begin{aligned}
& v=\frac{\lambda}{T}=\frac{\omega}{k}=\frac{100}{0.01} \quad\left[\begin{array}{l}
\text { As, } \omega=\frac{2 \pi}{T} \\
\text { and } k=\frac{2 \pi}{\lambda}
\end{array}\right] \\
& =10^4 \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
\)

Hint

\(f=\frac{1}{2 l}\left[\frac{T}{\mu}\right]^{\frac{1}{2}}\). When \(f\) is halved, the length is doubled.

Explanation:

\(
\begin{array}{ll}
\therefore & \frac{f_1}{f_2}=\frac{l_2}{l_1} \\
\text { Here, } f_1=512 \mathrm{~Hz}, {l}_1=0.5 \mathrm{~m} \\
& f_2=256 \mathrm{~Hz}, {l}_2=? \\
\therefore & \frac{512}{256}=\frac{l_2}{0.5} \\
\Rightarrow & l_2=0.5 \times 2=1 \mathrm{~m}
\end{array}
\)

Hint

Beat period (T) is the time interval between two successive beats.
\(
T=\frac{1}{\text { Beat frequency }}=\frac{1}{f_1-f_2}
\)
Frequency of beats will be \(\frac{1}{3}-\frac{1}{4}=\frac{1}{12}\). Hence time period \(=12 \mathrm{~s}\).

Hint

We have \(c \propto \sqrt{T}\)
\(
\begin{aligned}
& \therefore \frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}} \\
& \Rightarrow \frac{c_1}{2 c_2}=\sqrt{\frac{(27+273) k}{T_2}} \\
& \Rightarrow T_2=1200 k=927^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

Given \(y=0.40 \cos [2000 t+0.80]\) Comparing with the standard equation \(y=\mathrm{A} \cos (2 \pi f t+\phi)\)
We get angular frequency, \(2 \pi f=2000\)
\(
f \Rightarrow \frac{2000}{2 \pi}=\frac{1000}{\pi} \mathrm{Hz}
\)

Hint

With the propagation of a longitudinal wave, energy alone is propagated. Energy is propagated along with the wave motion without any net transport of the matter.

Hint

For the production of beats, different frequencies are essential. The different amplitudes affect the minimum and maximum amplitude of the beats and different phases affect the time of occurrence of minimum and maximum.

Hint

In a closed organ pipe, only alternate harmonics of frequencies \(f_1, 3 f_1, 5 f_1, \ldots\) etc are present. The harmonics of frequencies \(2 f_1, 4 f_1, 6 f_1, \ldots\) are missing. In general, the frequency of note produced in nth normal mode of vibration of a closed organ pipe would be
\(
f_n=\frac{(2 n-1) v}{4 L}=(2 n-1) f_1
\)
This is \((2 n-1)\) th harmonic or \((n-1)\) th overtone.
Third overtone has a frequency \(7 f_1\), which means
\(
L=\frac{7 \lambda}{4}=\frac{\lambda}{2}+\frac{\lambda}{2}+\frac{\lambda}{2}+\frac{\lambda}{4}
\)
which is three full loops and a half loop, which is equal to four nodes and four antinodes.

Hint

Here, \(f^{\prime}=\frac{9}{8} f\)
Source and observer are moving in opposite direction, therefore, apparent frequency
\(
\begin{aligned}
& f^{\prime}=f \times \frac{(v+u)}{(v-u)} \\
& \frac{9}{8} f=f \times\left(\frac{340+u}{340-u}\right) \\
& \Rightarrow 9 \times 340-9 u=8 \times 340+8 u \\
& \Rightarrow 17 u=340 \times 1 \Rightarrow u=\frac{340}{17}=20 \mathrm{~m} / \mathrm{sec}
\end{aligned}
\)

Hint

Given \(y=4 \sin \left(\frac{\pi}{6}\right) \sin (3 x-15 t)\)
Compare the given equation with standard form
\(
y=A \sin \left[\frac{2 \pi x}{\lambda}-\frac{2 \pi t}{T}\right]
\)
where, \(\frac{2 \pi}{\lambda}=3, \lambda=\frac{2 \pi}{3}\) and \(\frac{2 \pi}{T}=15\)
\(
T=\frac{2 \pi}{15}
\)
Speed of propagation,
\(
v=\frac{\lambda}{T}=\frac{2 \pi / 3}{2 \pi / 15}=5
\)

Hint

At a given time \((t=\) constant \()\), the phase changes with position \(x\).
The phase change \((\Delta \phi)\) at a given time for a wavelength \((\lambda)\) for a distance \(\Delta x\) is given by
\(
\Delta \phi=\frac{2 \pi}{\lambda} \Delta x \dots(i)
\)
From Eq. (i), \(\Delta x=\frac{\lambda}{2 \pi} \cdot \Delta \phi\) or \(\lambda=2 \pi \cdot \frac{\Delta x}{\Delta \phi}\)
Here, \(\quad \Delta x=0.4 \mathrm{~m}\)
\(
\begin{aligned}
\Delta \phi & =1.6 \pi \\
\therefore \quad \lambda & =2 \pi \cdot \frac{0.4}{1.6 \pi}=0.5
\end{aligned}
\)
\(\therefore\) Frequency of wave is
\(
f=\frac{v}{\lambda}=\frac{330}{0.5}=660 \mathrm{~Hz}
\)
where, \(v=330 \mathrm{~m} / \mathrm{s}=\) velocity of sound

Hint

The velocity of propagation of a transverse wave on a stretched string is given by
\(
v=\sqrt{\left(\frac{T}{\mu}\right)}
\)
where \(T\) is tension in the string and \(\mu\) is linear density of the string i.e. mass per unit length of the string.
Here, \(\quad \mu=\frac{0.035}{5.5} \mathrm{~kg} / \mathrm{m}\)
\(
\begin{aligned}
& T=77 \mathrm{~N} \\
\therefore \quad v & =\sqrt{\frac{77 \times 5.5}{0.035}}=110 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

Intensity is proportional to (amplitude) \({ }^2\) and also intensity \(\alpha\) (frequency) \({ }^2\). Therefore, intensity becomes \(\frac{2^2}{4^2}=\frac{1}{4}\) th

Hint

The frequency of a sound wave remains independent of the medium through which it travels. Wavelength, velocity, and phase change with the change in media.

Hint

From purely theoretical considerations, Newton came to the conclusion that velocity of longitudinal waves through any medium; solid, liquid or gas depends upon the elasticity and density of the medium. Newton gave the formula
\(
v=\sqrt{\left(\frac{E}{\rho}\right)}
\)
where \(v=\) velocity of sound in the medium \(E=\) coefficient of elasticity in the medium
\(\rho=\) density of the (undisturbed) medium

Hint

Equation of plane progressive simple harmonic wave is
\(
y=a \sin \left[2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)+\phi\right] \dots(i)
\)
The given equation is
\(
y=4 \sin \left[\pi\left(\frac{t}{5}-\frac{x}{9}\right)+\frac{\pi}{6}\right]
\)
Multiplying and dividing \(\left(\frac{t}{T}-\frac{x}{\lambda}\right)\) by 2 .
It is written as,
\(
y=4 \sin \left[2 \pi\left(\frac{t}{10}-\frac{x}{18}\right)+\frac{\pi}{6}\right] \dots(ii)
\)
Comparing Eqs. (i) and (ii), we find \(a=4 \mathrm{~cm}, \quad T=10 \mathrm{~s}, \quad \lambda=18 \mathrm{~cm}\) and
\(
\phi=\frac{\pi}{6}
\)

Hint

de-Broglie wavelength associated with a particle is given by
\(
\begin{aligned}
& \lambda=\frac{h}{p} \\
& \lambda \propto \frac{1}{p}
\end{aligned}
\)