Waves

$C_1=\sqrt{\frac{T}{\mu}}, C_2=\sqrt{\frac{T}{4 \mu}}=\frac{C_1}{2}$

For node at O :

$\begin{aligned} & \mathrm{L}=\frac{\mathrm{n} \lambda_1}{2}, 2 \mathrm{~L}=\frac{\mathrm{m} \lambda_2}{2}(\mathrm{n}, \mathrm{m} \text { are integers }) \\ & \lambda_1=\frac{2 \mathrm{~L}}{\mathrm{n}}, \lambda_2=\frac{4 \mathrm{~L}}{\mathrm{~m}} \\ & \frac{\mathrm{C}_1}{\lambda_1}=\frac{\mathrm{C}_2}{\lambda_2} \\ & \Rightarrow \frac{\mathrm{C}_1}{\frac{2 \mathrm{~L}}{\mathrm{n}}}=\frac{\frac{\mathrm{C}_1}{2}}{\frac{2 \mathrm{~L}}{\mathrm{~m}}} \\ & \Rightarrow 4 \mathrm{n}=\mathrm{m} \end{aligned}$

For minimum frequency, $\mathrm{n}=1, \mathrm{~m}=4$

$\therefore v_{\min }=\frac{\mathrm{C}_1 \times 1}{2 \mathrm{~L}}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}=v_0$

The string will look like

Total no. of nodes $=6$ including the end nodes

For antinode at O :

$\begin{aligned} & \mathrm{L}=(2 \mathrm{n}+1) \frac{\lambda_1}{4} ; 2 \mathrm{~L}=(2 \mathrm{n}+1) \frac{\lambda_2}{4} \quad \text { ( } \mathrm{n}, \mathrm{m} \text { are integers) } \\ & \lambda_1=\frac{4 \mathrm{~L}}{(2 \mathrm{n}+1)} ; \lambda_2=\frac{8 \mathrm{~L}}{(2 \mathrm{~m}+1)} \\ & \frac{C_1}{\lambda_1}=\frac{C_2}{\lambda_2} \\ & \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{\lambda_1}{\lambda_2} \\ & 2=\frac{\frac{4 L}{(2 n+1)}}{\frac{8 \mathrm{~L}}{(2 m+1)}} \\ \end{aligned}$

$4=\frac{(2 m+1)}{(2 n+1)} \Rightarrow \text { even }=\frac{\text { odd }}{\text { odd }} \Rightarrow \text { This node is not possible }$

Natural frequency of closed pipe,

f = (2n + 1)f₀

f₀ is fundamental frequency

n = 0, 1, 2 .......


Frequency of source received by pipe,

$f' = {f_s}\left[ {{{v - 0} \over {v - u}}} \right]$

For resonance,

$f' = f$

${f_s}\left[ {{v \over {v - u}}} \right] = (2n + 1){f_0}$

If u = 0.8v, f_s = f₀

$f' = {v \over {0.2v}}{f_0} = 5{f_0}$

For n = 2 pipe can be in resonance

Hence, option (a) is correct.

If u = 0.8v, f_s = 2f₀

$f' = {v \over {0.2v}} \times 2{f_0} = 10{f_0}$

If u = 0.8v, f_s = 0.5f₀

$f' = {v \over {0.2v}} \times 0.5{f_0} = 2.5{f_0}$

Not possible.

If u = 0.5v, f_s = 1.5f₀

$f' = {v \over {0.5v}} \times 1.5{f_0} = 3{f_0}$

For n = 1 f = 3f₀

Pipe can be in resonance.

Hence, option (d) is correct.

To determine which statements are true, we'll analyze the experiment step by step, using the theory of resonating air columns in a closed tube (closed at one end and open at the other).

Fundamentals of Resonance in a Closed Tube

For a tube closed at one end, the resonant lengths occur at:

$ L_n = \left( \frac{2n - 1}{4} \right) \lambda + e $

where:

$ L_n $ is the length of the air column for the $ n^\text{th} $ resonance,

$ \lambda $ is the wavelength of the sound wave,

$ e $ is the end correction,

$ n = 1, 2, 3, \dots $

Successive resonances (like those at $ L_1 $ and $ L_2 $) can be used to eliminate the end correction and find the wavelength.

Calculations

Step 1: Calculate the Wavelength ($ \lambda $)

Given:

First resonance length ($ L_1 $) = 50.7 cm

Second resonance length ($ L_2 $) = 83.9 cm

The difference between successive resonances in a closed tube is:

$ L_2 - L_1 = \frac{\lambda}{2} $

Solving for $ \lambda $:

$ \lambda = 2(L_2 - L_1) = 2(83.9\, \text{cm} - 50.7\, \text{cm}) = 2(33.2\, \text{cm}) = 66.4\, \text{cm} $

Conclusion: The wavelength of the sound wave is 66.4 cm.

Option C is true.

Step 2: Calculate the Speed of Sound ($ v $)

Using the relationship $ v = f\lambda $:

Frequency ($ f $) = 500 Hz

Wavelength ($ \lambda $) = 66.4 cm = 0.664 m

$ v = f\lambda = 500\, \text{Hz} \times 0.664\, \text{m} = 332\, \text{m/s} $

Conclusion: The speed of sound determined from the experiment is 332 m/s.

Option A is true.

Step 3: Calculate the End Correction ($ e $)

Assuming the first resonance corresponds to the second harmonic ($ n = 2 $):

$ L_1 = \left( \frac{2n - 1}{4} \right) \lambda + e $

For $ n = 2 $:

$ L_1 = \left( \frac{3}{4} \right) \lambda + e $

Solving for $ e $:

$ e = L_1 - \left( \frac{3}{4} \right) \lambda = 50.7\, \text{cm} - \left( \frac{3}{4} \times 66.4\, \text{cm} \right) $

$ e = 50.7\, \text{cm} - 49.8\, \text{cm} = 0.9\, \text{cm} $

Conclusion: The end correction is 0.9 cm.

Option B is true.

Step 4: Determine if $ L_1 $ is the Fundamental Harmonic

Assuming $ L_1 $ corresponds to the fundamental harmonic ($ n = 1 $):

$ L_1 = \left( \frac{1}{4} \right) \lambda + e $

$ e = L_1 - \left( \frac{1}{4} \right) \lambda = 50.7\, \text{cm} - 16.6\, \text{cm} = 34.1\, \text{cm} $

An end correction of 34.1 cm is unreasonably large for typical experiments, indicating that $ L_1 $ does not correspond to the fundamental harmonic.

Conclusion: The resonance at 50.7 cm does not correspond to the fundamental harmonic.

Option D is false.

The velocity of the transverse wave is given by

$v = \sqrt {{T \over \mu }} $ ..... (i)

Where T is the tension in the rope and is given by

T = Mg + $\mu$yg

Where y = distance of point P on the rope from its bottom

L = total length of the rope

Put the value of T in eqn. (i), we get

$v = \sqrt {{{Mg + \mu yg} \over \mu }} $ ...... (ii)

Also, speed of pulse 1 and pulse 2 are same at each point because speed is a characteristic of the medium. Hence, speed of the two pulses are same at the mid point (x = l/2). However, since pulse 1 is moving in +x direction and pulse 2 is moving in −x direction, their velocities (which is a vector quantity) at the mid point are not same.

So, option (b) is incorrect.

Since both the pulse travel in the same medium and velocity is medium dependent. The velocity of any pulse along the rope depends on tension in the rope and linear mass density so it is independent of its frequency(characteristic of the source of pulse generation) and wavelength.

So, option (d) is correct.

Also, v = $\nu $$\lambda$ or $\lambda$ $\propto$ v

($\because$ frequency of a wave depends on the source)

For pulse 1, the velocity decreases at point A as tension decreases. So the wavelength of pulse 1 becomes shorter when it reaches point A.

So, option (c) is incorrect.

From eqn. (ii), the velocity is given as

$v = \sqrt {{{Mg + \mu yg} \over \mu }} $ or ${{dy} \over {dt}} = \sqrt {{{Mg + \mu yg} \over \mu }} \Rightarrow dt = {{dy} \over {\sqrt {{{Mg + \mu yg} \over \mu }} }}$

Integrating both sides,

$\int\limits_0^T {dt = \int\limits_0^L {{{dy} \over {\sqrt {{{Mg} \over \mu } + yg} }}} } $; $\therefore$ $T = {2 \over g}\left( {\sqrt {{{Mg} \over \mu } + Lg} - \sqrt {{{Mg} \over \mu }} } \right)$

Where T is time taken by pulse to travel distance L.

Since T depends upon g, M, L and $\mu$ which are constant.

$\therefore$ Time taken by pulse 1 to reach point A and time taken by pulse 2 to reach point O is same.

$\therefore$ T_AO = T_OA; So option (a) is correct.

Consider the time $t\left(\leq t_Q\right)$ when car is at $S_1$ between $P$ and $Q$. The distance travelled by the car in time $t$ is $\mathrm{PS}_1=u t$. At this instant, the lines $\mathrm{S}_1{N}$ and $\mathrm{S}_1 \mathrm{M}$ both make angle $\theta$ with the velocity vector $\vec{u}$. The component of observer (person sitting in the car) velocity towards the sources $N$ and $M$ is $u_o=u \cos \theta$. The sources $N$ and $M$ are at rest i.e., $u_s=0$. Apply Doppler's effect equation to get frequencies of the sources $N$ and $M$ heard by the observer as

$ \begin{aligned} \nu_N^{\prime} & =\frac{v+u_o}{v-u_s} \nu_N=\frac{v+u \cos \theta}{v} \nu_N \\\\ & =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}\right) \nu_N, \\\\ \nu_M^{\prime} & =\frac{v+u_o}{v-u_s} \nu_M=\frac{v+u \cos \theta}{v} \nu_M \\\\ & =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}\right) \nu_M . \end{aligned} $

The beat frequency heard by the observer at time $t(\leq$ $\left.t_Q\right)$ is

$ \begin{aligned} \nu(t) & =\nu_N^{\prime}-\nu_M^{\prime} \\\\ & =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}\right)\left(\nu_N-\nu_M\right) ........(1) \end{aligned} $

Now, consider the time $t\left(\geq t_Q\right)$ when car is at $S_2$ between $Q$ and $R$. The distance travelled by the car in time $t$ is $\mathrm{PS}_2=u t$. At this instant, the lines $\mathrm{S}_2 \mathrm{~N}$ and $\mathrm{S}_2 \mathrm{M}$ both make angle $\left(180^{\circ}-\theta\right)$ with the velocity vector $\vec{u}$. The component of observer velocity towards the sources $N$ and $M$ is $u_o=-u \cos \theta$. Apply Doppler's effect equation to get

$ \begin{aligned} \nu_N^{\prime} & =\frac{v+u_o}{v-u_s} \nu_N=\frac{v-u \cos \theta}{v} \nu_N \\\\ & =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}\right) \nu_N, \\\\ \nu_M^{\prime} & =\frac{v+u_o}{v-u_s} \nu_M=\frac{v-u \cos \theta}{v} \nu_M \\\\ & =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}\right) \nu_M . \end{aligned} $

The beat frequency heard by the observer at time $t(\geq$ $\left.t_Q\right)$ is

$ \begin{aligned} \nu(t) & =\nu_N^{\prime}-\nu_M^{\prime} \\\\ & =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}\right)\left(\nu_N-\nu_M\right) ..........(2) \end{aligned} $

Substitute $t=0$ and $t=t_Q={D \over u}$ in equation (1) to get $\nu_P$ and $\nu_Q$ and substitute $t=t_R=2 {D \over u}$ in equation (2) to get $\nu_R$ i.e.,

$ \begin{aligned} & \nu_P=\nu(t=0)=\left(1+\frac{u}{v} \frac{D}{\sqrt{d^2+D^2}}\right)\left(\nu_N-\nu_M\right) \\\\ & \nu_Q=\nu(t=D / u)=\left(\nu_N-\nu_M\right) \\\\ & \nu_R=\nu(t=2 D / u)=\left(1-\frac{u}{v} \frac{D}{\sqrt{d^2+D^2}}\right)\left(\nu_N-\nu_M\right) \end{aligned} $

$ \therefore $ $ \nu_P+\nu_R=2 \nu_Q $

Differentiate equations (1) w.r.t. time t to get rate of change of beat frequency

$ \frac{\mathrm{d} \nu(t)}{\mathrm{d} t}=-\left(\nu_N-\nu_M\right) \frac{u^2}{v} \frac{d^2}{\left(d^2+(D-u t)^2\right)^{3 / 2}} $ ...........(3)

From equation (3), the slope is negative and its magnitude is maximum when $t={D \over u}=t_Q$ (denominator is minimum). Thus the rate of change of beat frequency is maximum when car passes through $Q$.

The displacement of a stationary wave is given by

$y(x,t) = A\sin \left( {{{2\pi x} \over \lambda }} \right)\cos (2\pi vt)$.

The boundary conditions give node at x = 0 and anti-node at x = 3 m i.e.,

y(0, t) = 0, ....... (1)

y(3, t) = $\pm$A. ..... (2)


The fundamental frequency is given by

${v_0} = {v \over \lambda } = {v \over {4l}} = {{100} \over {4(3)}} = {{25} \over 3}$ Hz.

Thus, the waveform will satisfy equations (1) and (2), and the permissible frequencies will be odd multiples of v₀.

Minimum length $ = {\lambda \over 4} \Rightarrow \lambda = 4l$

Now, v = f $\lambda$ = (244) $\times$ 4 $\times$ l

as l = 0.350 $\pm$ 0.005

$\Rightarrow$ v lies between 336.7 m/s to 346.5 m/s

Now, $v = \sqrt {{{\gamma RT} \over {M \times {{10}^{ - 3}}}}} $, here M is molecular mass in gram $ = \sqrt {100\gamma RT} \times \sqrt {{{10} \over m}} $.

For monoatomic gas,

$\gamma$ = 1.67

$\Rightarrow$ $v = 640 \times \sqrt {{{10} \over m}} $

For diatomic gas,

$\gamma = 1.4 \Rightarrow v = 590 \times \sqrt {{{10} \over m}} $

$\therefore$ ${v_{Ne}} = 640 \times {7 \over {10}} = 448$ m/s

${v_{Ar}} = 640 \times {{17} \over {32}} = 340$ m/s

${v_{{O_2}}} = 590 \times {9 \over {16}}=331.8$ m/s

${v_{{N_2}}} = 590 \times {3 \over 5} = 354$ m/s

$\therefore$ Only possible answer is Argon.

Doppler's effect gives

${f_2} = \left[ {{{v + {u_0}} \over {v - {u_s}}}} \right]{f_1}$ ..... (1)

Let the wind speed be w and wind moves towards the source in case (i) and towards the observer in case (ii) (see figure).

In the case (i), u₀, the speed of observer w.r.t. medium considered positive when it moves towards the source is +(u $-$ w) and u_s, the speed of source w.r.t. medium considered positive when it moves towards observer is +(u + w). Substitute these values in equation (1) to get

${f_2} = \left[ {{{v + (u - w)} \over {v - (u + w)}}} \right]{f_1}$,

which is greater than f₁ (assuming u > w). In the case (ii), u₀ is +(u + w) and u_s is +(u $-$ w). Substitute these values in equation (1) to get

${f_2} = \left[ {{{v + (u + w)} \over {v - (u - w)}}} \right]f1$,

which is again greater than f₁.

Number of nodes = 6

From the given equation, we can see that

$k = {{2\pi } \over \lambda } = 62.8$ m^$-$1

$\therefore$ $\lambda = {{2\pi } \over {62.8}}$ m = 0.1 m

$1 = {{5\lambda } \over 2}$ = 0.25 m

The mid-point of the string is P, an antinode

$\therefore$ maximum displacement = 0.01 m

$\omega$ = 2$\pi$f = 628 s^$-$1

$\therefore$ $f = {{628} \over {2\pi }} = 100$ Hz

But this is fifth harmonic frequency.

$\therefore$ Fundamental frequency f₀

$ = {f \over 5} = 20$ Hz.

We have,

$\overrightarrow J = \overrightarrow r \times \overrightarrow F $

$|\overrightarrow J | = rF\sin 180^\circ = 0$

Therefore, the angular momentum is conserved, that is,

$mvr\sin \theta = $ Constant

where $\theta$ is the angle between the velocity $\overrightarrow v $ and position vector $\overrightarrow r $ of $-$q with respect to Q.

Fundamental frequency is maximum when length is minimum i.e. L₀,

Case 1. $L = {L_0},\,T = {T_0},\,f = {f_0}$;

${f_1} = {1 \over {2{L_0}}}\sqrt {{{{T_0}} \over \mu }} $

Case 2. ${f_2} = {1 \over {{L_0}}}\sqrt {{{{T_2}} \over {2\mu }}} = {{{f_0}} \over {\sqrt 2 }}$

Case 3. ${f_3} = {1 \over {{L_0}}}\sqrt {{{{T_2}} \over {3\mu }}} = {{{f_0}} \over {\sqrt 3 }}$

Case 4. ${f_4} = {1 \over {{L_0}}}\sqrt {{{{T_2}} \over {4\mu }}} = {{{f_0}} \over 2}$

Case 1. $L = {L_0},\,T = {T_0},\,f = {f_0}$

${f_1} = {1 \over {2{L_0}}}\sqrt {{{{T_0}} \over \mu }} $

Case 2. $L = {{3{L_0}} \over 2}$

${f_2} = {3 \over {2 \times {{3{L_0}} \over 2}}}\sqrt {{{{T_2}} \over {2\mu }}} = {f_0}$

$ \Rightarrow {f_0} = {1 \over {2{L_0}}}\sqrt {{{{T_2}} \over \mu }} \Rightarrow {T_2} = {{{T_0}} \over 2}$

Case 3. $L = {{5{L_0}} \over 4}$

${f_3} = {5 \over {2 \times {{5{L_0}} \over 4}}}\sqrt {{{{T_3}} \over {3\mu }}} = {f_0}$

$ \Rightarrow {f_0} = {2 \over {\sqrt 3 {L_0}}}\sqrt {{{{T_3}} \over \mu }} \Rightarrow {T_3} = {{3{T_0}} \over {16}}$

Case 4. $L = {{7{L_0}} \over 4}$

${f_4} = {{14} \over {2 \times {{7{L_0}} \over 4}}}\sqrt {{{{T_4}} \over {4\mu }}} = {f_0}$

$ \Rightarrow {f_0} = {2 \over {{L_0}}}\sqrt {{{{T_4}} \over \mu }} \Rightarrow {T_4} = {{{T_0}} \over {16}}$

A student is conducting a resonance column experiment. The tube has a diameter of 4 cm, and the tuning fork vibrates at 512 Hz. The air temperature is 38^oC, where the speed of sound is 336 m/s. The zero mark on the meter stick aligns with the top of the tube. For the first resonance, the water level reads :

During the first resonance :

$ \frac{\lambda}{4} = l_1 + e $

where $ l_1 $ is the length of the air column at resonance

Given that :

$ \text{end correction} = e = 0.6r = 0.6 \times 2 = 1.2 \, \text{cm} $

where $ r $ is the radius of the tube

The wavelength $ \lambda $ can be expressed as :

$ \lambda = 4(l_1 + e) $

The frequency $ f $ is related to the wavelength $ \lambda $ by :

$ f = \frac{v}{\lambda} = \frac{v}{4(l_1 + e)} $

Rearranging for $ l_1 $ :

$ 4(l_1 + e) = \frac{v}{f} $

Solving for $ l_1 $ :

$ l_1 = \frac{v}{4f} - e $

Substitute the given values :

$ l_1 = \frac{336 \, \text{m/s}}{4 \times 512 \, \text{Hz}} - 0.012 \, \text{m} $

$ l_1 = \frac{336}{2048} - 0.012 $

$ l_1 = 0.164 \, \text{m} - 0.012 \, \text{m} $

$ l_1 = 0.152 \, \text{m} $

Converting to cm :

$ l_1 = 15.2 \, \text{cm} $

u = 36 km h^$-$1 = 10 ms^$-$1,

$v$ = 320 ms^$-$1, $\nu $ = 8 kHz

The sound reflected from the building may be imagined to be coming from the mirror image. The driver is approaching the image-source which is also approaching him with the same speed. Hence the frequency of sound heard by the driver is

$\nu $' = $\nu $ $\left( {{{v + u} \over {v - u}}} \right)$

= 8 kHz $\times$ $\left( {{{320 + 10} \over {320 - 10}}} \right)$ = 8.5 kHz

(1) For a pipe closed at one end, we have

${{{\lambda _f}} \over 4} = L$ or ${\lambda _f} = 4L$

Therefore, the sound waves are longitudinal.

(2) For a pipe open at both ends, we have

${{{\lambda _f}} \over 2} = L$ or ${\lambda _f} = 2L$

Therefore, the sound waves are longitudinal.

(3) For a stretched wire clamped at both ends, we have

${{{\lambda _f}} \over 2} = L$

Vibration on the string is transverse.

(4) For a stretched wire clamped at both ends and at mid-point, we have

${{{\lambda _f}} \over 2} = {L \over 2} \Rightarrow {\lambda _f} = L$

Therefore, the vibration on the string is transverse.

The fundamental mode in a pipe closed at one end and the second harmonic in a string are shown in the figure.

Fundamental frequency of a hollow pipe closed at one end is

${\upsilon _p} = {v \over {4L}}$

where, v = speed of sound, L = length of a pipe

Frequency of second harmonic of a string is

${\upsilon _s} = {2 \over {2l}}\sqrt {{T \over \mu }} = {1 \over l}\sqrt {{T \over \mu }} $

where, T = tension of the string, m = mass per unit length of the string, l = length of the string

${\upsilon _s} = {1 \over l}\sqrt {{T \over {{m \over l}}}} = {1 \over l}\sqrt {{{Tl} \over m}} = \sqrt {{T \over {ml}}} $ ...... (i)

where m is the mass of the string

According to given problem, ${\upsilon _p} = {\upsilon _s}$

$\therefore$ ${v \over {4L}} = \sqrt {{T \over {ml}}} $ or $m = {{16{L^2}T} \over {{v^2}l}}$

Substituting the given values, we get

$m = {{16 \times {{(0.8)}^2} \times (50)} \over {{{(320)}^2} \times 0.5}} = 0.01$ kg = 10 gram

Velocity of point $p=dy/dt$

= $-$(velocity of wave $\times$ $dy/dx$)

Here, $dy/dx=$ negative ($-$ve). Therefore, velocity at point P is positive and is along X-axis only.

Equation of a wave moving in positive X-axis is given as,

$y = A\sin (wt - \phi )$ .... (i)

or, ${V_P} = Aw\cos (wt - \phi )$ .... (ii)

Here, $y' = 5$ cm, $A = 10$ cm

$\therefore$ $5 = 10\sin (wt - \phi ) \Rightarrow wt - \phi = 30^\circ $ .... (iii)

Put equation (iii) in equation (ii),

${V_P} = 0.10 \times w\cos 30^\circ $

Now, $V = {V_\lambda }$, $\therefore$ $v = V/\lambda = {{0.10} \over {0.5}} = 0.2$

$\therefore$ $w = 2\pi v = 2\pi \times 0.2 = 0.4\pi $

$ \Rightarrow {V_P} = 0.1 \times 0.4\pi \times {{\sqrt 3 } \over 2} = {{\sqrt 3 } \over {50}}\pi \,\widehat j$ m/s

It is in positive Y-direction.

Let f be frequency of string

According to close open pipe, we have

$ \Rightarrow f = {{3V} \over {4l}} = {{3 \times 340} \over {4 \times 75 \times {{10}^{ - 2}}}}$

$ = 340$ Hz

Where, $V = 340$ m/s

$l = 75$ cm $ \times {10^{ - 2}}$ m

$ = 75 \times {10^{ - 2}}$ m

And, n = frequency of tuning fork

$ = (340 + 4)$ or $(340 - 4)$

$ \Rightarrow n = 344$ or 336

As tension T increase, beat frequency decrease

But frequency of tuning fork increases

$\therefore$ $n = 344$ Hz

Wave produced by tuning Fork in a vertical plane as shown in above figure.

Therefore, fringes of tuning Fork is kept in vertical plane.

$v_{\mathrm{SA}}=(340+20) \mathrm{m} / \mathrm{s}=360 \mathrm{~m} / \mathrm{s} $

$v_{\mathrm{SB}}=(340-30) \mathrm{m} / \mathrm{s}=310 \mathrm{~m} / \mathrm{s} $

where, speed of sound in air = 340 m/s

Hence, the speed of sound depends on the frame of reference of observer.

Since all the passengers in train A are moving with a velocity of $20 \mathrm{~m} / \mathrm{s}$ therefore the distribution of sound intensity of the whistle by the passengers in the train $A$ is uniform.

Trick : Total frequency $=(1120-800) \mathrm{Hz} =320 \mathrm{~Hz}$

There is no relative motion between source and observer.

Therefore, all the passengers will hear the sound of same intensity.

Frequency $(f') = {f_1}\left( {{{v - {v_0}} \over {v - {v_s}}}} \right) = 800\left[ {{{340 - 30} \over {340 - 20}}} \right]$

$=800 \times \frac{31}{32}$ ..... (i)

where, $\left(\frac{v-v_{0}}{v-v_{\mathrm{S}}}\right)=\frac{31}{32}$ ..... (ii)

Frequency $\left(f^{\prime}\right)=f_{2}\left(\frac{v-v_{0}}{v-v_{\mathrm{S}}}\right)=1120 \times \frac{31}{32}$ ..... (iii)

$\therefore \quad f^{\prime}-f^{\prime}=(1120-800) \frac{31}{32}=320 \times \frac{31}{32}$

$=310 \mathrm{~Hz}$

Reason:

In above figure shown that S.H.M

Therefore, $f_s=Kx$

Given, $v = {c_1}\sqrt {{c_2} - {x^2}} $ ..... (i)

For a simple harmonic motion,

$v = \omega \sqrt {{A^2} - {x^2}} $ ..... (ii)

On comparing eq. (i) & (ii) therefore it shows the S.H.M of particle.

A $\to$ P

(b) Given,

$v=-kx$

At $x=0,v=0$

$x=2,v=-2k$

$x=-5,v=+5k$

The above consideration show that particle is coming from minus x-axis.

Velocity is decreasing with time.

Graph:

Therefore, no change in direction, velocity keeps on decreasing. So that kinetic energy is decreasing with time.

B $\to q_1r$

(C) From the reference frame of the observer in the elevator a constant force acts on the object. Hence the motion of the object will remain simple harmonic.

Therefore, option 1 is correct.

Velocity of object

$v = 2\sqrt {G{{{M_e}} \over {{R_e}}}} = \sqrt 2 \sqrt {2G{{{M_e}} \over {{R_e}}}} = \sqrt 2 {v_e}$

Since, $v > {v_e}$ (where $v_e$ is escape velocity) object will move out of earth.

Hence, option (Q) and (R) are correct.

Frequency of $1^{\text {st }}$ harmonic of $\mathrm{AB}=\frac{1}{2 l} \sqrt{\frac{\mathrm{~T}_{\mathrm{AB}}}{m}}$

Frequency of $2^{\text {nd }}$ harmonic of $\mathrm{CD}=\frac{1}{l} \sqrt{\frac{\mathrm{~T}_{\mathrm{CD}}}{m}}$

Given that 2 frequencies are equal.

$ \begin{aligned} \frac{1}{2 l} \sqrt{\frac{\mathrm{~T}_{\mathrm{AB}}}{m}} & =\frac{1}{l} \sqrt{\frac{\mathrm{~T}_{\mathrm{CD}}}{m}} \\ \frac{\mathrm{~T}_{\mathrm{AB}}}{m} \times \frac{1}{4 l^2} & =\frac{1}{l^2} \cdot \frac{\mathrm{~T}_{\mathrm{CD}}}{m} \quad(\text { On squaring }) \\ \frac{\mathrm{T}_{\mathrm{AB}}}{4} & =\mathrm{T}_{\mathrm{CD}} \\ \mathrm{~T}_{\mathrm{AB}} & =4 \mathrm{~T}_{\mathrm{CD}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

For rotational equilibrium of massless rod, Taking torque about point D ,

$ \mathrm{T}_{\mathrm{AB}} \times x=\mathrm{T}_{\mathrm{CD}}(\mathrm{~L}-x) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

For translational equillibrium,

$ \mathrm{T}_{\mathrm{AB}}+\mathrm{T}_{\mathrm{CD}}=m g \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

From (i) and (iii), $4 \mathrm{~T}_{\mathrm{CD}}+\mathrm{T}_{\mathrm{CO}}=m g$

$ \begin{array}{ll} \Rightarrow & \mathrm{T}_{\mathrm{CD}}=\frac{m g}{5} \\ \therefore & \mathrm{~T}_{\mathrm{AB}}=4 \times \frac{m g}{5}=\frac{4 m g}{5} \end{array} $

On substituting above value in (i) we get

$ \Rightarrow \quad \begin{aligned} \frac{4 m g}{5} \times x & =\frac{m g}{5}(L-x) \\ 4 x & =\mathrm{L}-x \\ 5 x & =\mathrm{L} \\ x & =\mathrm{L} / 5 . \end{aligned} $

Number of times the intensity is maximum on superposition in 1 , is calculated as follows:

Here,

$ \begin{aligned} y_1 & =\mathrm{A} \cos (0.5 \pi x-100 \pi t) \\ y_1 & =\mathrm{A} \cos \left(\frac{2 \pi}{\lambda} x-2 \pi v_1 t\right) \\ y_2 & =\mathrm{A} \cos (0.46 \pi-92 \pi t) \\ & =\mathrm{A} \cos \frac{2 \pi}{\lambda^2} x-2 \pi v_2 t \end{aligned} $

The two waves travelling in same direction will form beats having frequency $=$ frequency of maximum

$ =\mathrm{v}_1=\mathrm{v}_2=50-46=4 \mathrm{~Hz} \text { or } 4 \mathrm{~s}^{-1} . $

$\left[\right.$ Here, $2 \pi v_1 t=100 \pi t$ and $2 \pi v_2 t=92 \pi, v_2=46 \mathrm{~Hz}$ ]

Comparing the louder wave with simple harmonic waves,

waves velocity is $\frac{w}{k}=\frac{100}{0.5}=200 \mathrm{~m} / \mathrm{s}$.

Average frequency of wave after superposition will be

$ f=\frac{f_1+f_2}{2}=\frac{100+92}{2}=96 \mathrm{~Hz} $

Let the speed of the train be V$_t$,

Let $v$ be the actual frequency of the whistle.

Speed of sound $\mathrm{V}_{s}=300 \mathrm{~m} / \mathrm{s}$

While train is approaching, $v_{2}=2.2 \mathrm{~kHz}$

$=2.2\times10^3$ Hz

$v_{2}=\frac{\mathrm{V}_{s}}{\mathrm{~V}_{s}-\mathrm{V}_{t}} \times v$

$2.2 \times 10^{3}=\frac{\mathrm{V}_{s}}{\mathrm{~V}_{s}-\mathrm{V}_{t}} \times v=\frac{300}{300-\mathrm{V}_{t}} \times v$ ...... (i)

While train is receding,

$v_1=1.8$ kHz = $1.8\times10^3$ Hz

${v_1} = {{{V_s}} \over {{V_s} + {V_t}}} \times v$

$1.8 \times {10^3} = {{300} \over {300 + {V_t}}} \times v$ .... (ii)

Dividing equation (i) by (ii)

${{2.2 \times {{10}^3}} \over {1.8 \times {{10}^3}}} = {{{{300} \over {300 - {V_t}}} \times v} \over {{{300} \over {300 + {V_t}}} \times v}}$

$ = {{2.2} \over {1.8}} = {{300 + {V_t}} \over {300 - {V_t}}}$

$660 - 2.2\,{V_t} = 540 + 1.8\,{V_t}$

$2.2\,{V_t} + 1.8\,{V_t} = 660 - 540$

$4{V_t} = 120$

${V_t} = 30$ m/s

Step 1: Understanding Symbols

Let $\omega$ be the angular velocity and $A$ be the amplitude.

Step 2: Maximum Velocity

The highest speed up and down (maximum transverse velocity) is $V_{\max} = \omega A = 3~\mathrm{m/s}$.

Step 3: Maximum Acceleration

The highest acceleration up and down is $\omega^2 A = 90~\mathrm{m/s}^2$.

Step 4: Wave Velocity Relation

The speed that the wave itself travels is $\frac{\omega}{k} = 20~\mathrm{m/s}$, where $k$ is called the wave number.

Step 5: Find Angular Velocity $\omega$

To find $\omega$, divide the maximum acceleration equation by the maximum velocity equation:

$\frac{A \omega^2}{A \omega} = \frac{90}{3}$

So, $\omega = 30~\mathrm{rad/s}$.

Step 6: Find Amplitude $A$

From the maximum velocity equation: $A = \frac{3}{30} = 0.1~\mathrm{m}$.

Step 7: Find Wave Number $k$

We know $k = \frac{\omega}{v}$, where $v$ is the wave speed.

$k = \frac{30}{20} = 1.5~\mathrm{m}^{-1}$

Step 8: Write the Wave Equation

The wave can be written as: $y = A \sin(\omega t + k x + \phi)$, where $\phi$ is a starting phase (it can take any value).

So, the waveform is: $y = 0.1 \sin(30t + 1.5x \pm \phi)$