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.

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.