Waves

For a closed organ pipe, only odd harmonics are present. The frequencies are:

$f_n = \frac{nv}{4L_c}, \quad n = 1, 3, 5, 7, ...$

The fifth harmonic (n = 5) of the closed pipe:

$f_{closed} = \frac{5v}{4L_c}$

For an open organ pipe, all harmonics are present. The frequencies are:

$f_n = \frac{nv}{2L_o}, \quad n = 1, 2, 3, ...$

The first harmonic (n = 1) of the open pipe:

$f_{open} = \frac{v}{2L_o}$

Setting Frequencies Equal (Unison)

Since they are in unison:

$\frac{5v}{4L_c} = \frac{v}{2L_o}$

Solving for the Ratio

Cancelling $v$ from both sides:

$\frac{5}{4L_c} = \frac{1}{2L_o}$

Cross-multiplying:

$5 \times 2L_o = 4L_c \times 1$

$10L_o = 4L_c$

$\frac{L_c}{L_o} = \frac{10}{4} = \frac{5}{2}$

Finding x

Given that the ratio is $\frac{5}{x}$:

$\frac{5}{x} = \frac{5}{2}$

$\boxed{x = 2}$

The answer is Option B: 2.

The volume ( $V$ ) of a sphere of radius $R$ is given by the formula :

$ V=\frac{4}{3} \pi R^3 $ .......(i)

According to the problem, the volume increases by 8 times ( $\mathrm{V}^{\prime}=8 \mathrm{~V}$ ) :

$ 8 V=\frac{4}{3} \pi\left(R^{\prime}\right)^3 \ldots $(ii)

$\Rightarrow $ $ 8 \times \frac{4}{3} \pi R^3=\frac{4}{3} \pi\left(R^{\prime}\right)^3 \Rightarrow R^{\prime}=2 R $

The intensity (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 $R^{\prime}=2 R$ :

$ \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} $

Therefore, when the volume of the spherical detector increases by 8 times, the intensity decreases by 4 times. Hence, the correct option is (B).

For an open organ pipe (open at both ends), the standing waves formed have antinodes at both ends.

The frequency of the $\mathrm{n}^{\text {th }}$ harmonic $\left(\nu_{\mathrm{n}}\right)$ is an integer multiple of the fundamental frequency and is given by the formula :

$ \mathrm{v}_{\mathrm{n}}=\frac{\mathrm{nv}}{2 \mathrm{~L}} $

Where :

• n is the harmonic number (integer: $1,2,3 \ldots$ )

• $v$ is the velocity of sound in the medium

• L is the length of the open pipe

Using the formula for the $\mathrm{n}^{\text {th }}$ harmonic :

Frequency of the $3^{\text {rd }}$ harmonic $=v_3=\frac{3 v}{2 L}$

Frequency of the $6^{\text {th }}$ harmonic $=v_6=\frac{6 \mathrm{v}}{2 \mathrm{~L}}$

The difference between the frequencies of $6^{\text {th }}$ and $3^{\text {rd }}$ harmonic is 2200 Hz ,

$ v_6-v_3=2200 $

$\Rightarrow \frac{6 \mathrm{v}}{2 \mathrm{~L}}-\frac{3 \mathrm{v}}{2 \mathrm{~L}}=2200$

$\Rightarrow \frac{3 \mathrm{v}}{2 \mathrm{~L}}=2200$

$\Rightarrow \frac{3 \times 330}{2 L}=2200$

$\Rightarrow \frac{990}{2 \mathrm{~L}}=2200$

$ \Rightarrow L=\frac{990}{2200 \times 2} \Rightarrow L=0.225 \mathrm{~m}=225 \mathrm{~mm} $

Therefore, the length of the pipe is 225 mm. Hence, the correct option is (B).

When a string with linear mass density $\mu$ (mass per unit length) is under tension $T$, the speed $v$ of a transverse wave is given by :

$ v=\sqrt{\frac{T}{\mu}} $

So, the time taken to reach the joint is $t=\frac{\text { Length }}{\text { Velocity }}=\frac{L}{v}$

Substituting the velocity formula into the time equation :

$ \mathrm{t}=\mathrm{L} \sqrt{\frac{\mathrm{~T}}{\mu}}=\mathrm{L} \sqrt{\frac{\mu}{\mathrm{~T}}} $

The two strings are joined together and tied between two rigid supports, meaning the tension ( $T$ ) is the same throughout both strings $(T=500 \mathrm{~N})$.

For String A (time $\mathrm{t}_1$ ) :

The length of the string $A$ is $L_A=2.5 \mathrm{~m}$ and the linear density is $\mu_A=2 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ So, the time taken to reach the end of string A is,

$ \mathrm{t}_1=2.5 \sqrt{\frac{2 \times 10^{-4}}{500}} $

For String B (time $\mathrm{t}_2$ ) :

The length of the string $B$ is $L_B=1.5 \mathrm{~m}$ and the linear density is $\mu_B=4 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ So, the time taken to reach the end of string B is,

$ \mathrm{t}_2=1.5 \sqrt{\frac{4 \times 10^{-4}}{500}} $

So, the ratio is,

$ \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{2.5 \sqrt{\frac{2 \times 10^{-4}}{500}}}{1.5 \sqrt{\frac{4 \times 10^{-4}}{500}}} $

$\Rightarrow \frac{t_1}{t_2}=\frac{2.5 \sqrt{2}}{1.5 \sqrt{4}}=\frac{5}{3 \sqrt{2}}$

$\Rightarrow $ $ t_1 t_2 \approx 1.18 $

Therefore, the ratio of the travel times to be approximately 1.18 .

Hence, the correct option is (B).

When a sound wave enters a split-tube system at point S , it divides into two components that travel along separate geometric paths before interference at the detector $D$.

The two split waves interfere with each other at D . For the detector to record a maximum amplitude (constructive interference), the path difference $\Delta \mathrm{x}$ between the two waves must be an integer multiple of the wavelength $\lambda$ :

$ \Delta \mathrm{x}=\left|\mathrm{x}_1-\mathrm{x}_2\right|=\mathrm{n} \lambda \text { where } \mathrm{n}=0,1,2,3, \ldots $

The wavelength of the sound wave is given as $\lambda=0.29 \mathrm{~m}$

The sound waves splits symmetrically at $S$ into straight path directly from $S$ to $D$ (Path 1) and semi-circular tracks of radius 0.51 (path 2).

Path 1 length $\mathrm{x}_1=1$

Path 2 length is $\mathrm{x}_2=\pi(0.5 \mathrm{l})=\frac{\pi \mathrm{l}}{2}$

So, path difference is $\Delta \mathrm{x}=\frac{\pi \mathrm{l}}{2}-\mathrm{l}=\mathrm{l}\left(\frac{\pi}{2}-1\right)$

For maximum intensity,

$ l\left(\frac{\pi}{2}-1\right)=\lambda \Rightarrow l=\frac{\lambda}{\pi / 2-1} $

Substituting $\lambda=0.29 \mathrm{~m}$ and $\pi \approx 3.1416$ :

$ \mathrm{l}=\frac{0.29}{3.1416 / 2-1} \approx 0.51 \mathrm{~m} $

So, $(\mathrm{P}) \rightarrow(3)$

The sound waves splits symmetrically at $S$ into Path 1 and path 2.

Path 1 length $\mathrm{x}_1=0.5 \mathrm{l}+\mathrm{l}+0.5 \mathrm{l}=2 \mathrm{l}$

Path 2 length is $\mathrm{x}_2=\mathrm{l}$

So, path difference is $\Delta \mathrm{x}=2 \mathrm{l}-\mathrm{l}=\mathrm{l}$

For maximum intensity,

$ \begin{gathered} \mathrm{l}=\lambda \\ \mathrm{l}=0.29 \mathrm{~m} \end{gathered} $

So, $(\mathrm{Q}) \rightarrow(4)$

The sound wave splits at the source S .

Path 1 consists of a straight vertical section of length 1 and a semi-circular path with diameter $d=l \sqrt{2}$

Therefore, the total length of Path 1 is

$ x_1=1+\frac{\pi l \sqrt{2}}{2}=l\left(1+\frac{\pi}{\sqrt{2}}\right) $

Path 2 is a direct straight line from S to D, so the length of Path 2 is $x_2 = l$

The path difference ($\Delta x$) between the two waves arriving at detector D is:

$\Delta x = x_1 - x_2$

$\Delta x = l\left(1 + \frac{\pi}{\sqrt{2}}\right) - l$

$\Delta x = \frac{\pi l}{\sqrt{2}}$

For the detector to record a maximum intensity/amplitude,

$\Delta x = \lambda$

$\frac{\pi l}{\sqrt{2}} = \lambda$

$l = \frac{\sqrt{2}\lambda}{\pi} = \frac{\sqrt{2} \times 0.29}{\pi} \approx 0.13$

So, (R) $\rightarrow$ (5)

The sound waves splits into two distinct paths forming a triangle with a base side of length 1 .

Path 1 is along two segments of a triangle forming angles of $45^{\circ}$ and $105^{\circ}$ at the base vertices S and D .

The third angle of this triangle is :

$ \theta_{\text {vertex }}=180^{\circ}-\left(45^{\circ}+105^{\circ}\right)=180^{\circ}-150^{\circ}=30^{\circ} $

By the law of sines :

$ \frac{l}{\sin \left(105^{\circ}\right)}=\frac{a}{\sin \left(30^{\circ}\right)}=\frac{b}{\sin \left(45^{\circ}\right)} $

$a=\frac{l \times \sin (30)}{\sin \left(105^{\circ}\right)}=\frac{l \times \frac{1}{2}}{\cos 15^{\circ}}=0.517 l$

$ \mathrm{b}=\frac{\mathrm{l} \times \sin \left(45^{\circ}\right)}{\sin \left(105^{\circ}\right)}=\frac{\mathrm{l} \times \frac{1}{\sqrt{2}}}{\cos 15^{\circ}}=0.732 \mathrm{l} $

The total length of Path 1 is the sum of these two sides:

$ x_2=a+b=0.517 l+0.732 l=1.249 l $

Path 2 is directly along the straight base line from S to D . Thus, the distance is $\mathrm{x}_2=\mathrm{l}$ The path difference $\Delta \mathrm{x}$ between Path 2 and Path 1 is:

$ \Delta \mathrm{x}=\mathrm{x}_2-\mathrm{x}_1=1.249 \mathrm{l}-\mathrm{l}=0.249 \mathrm{l} $

For constructive interference at the detector,

$ 0.249 \mathrm{l}=0.29 \Rightarrow \mathrm{l}=\frac{0.29}{0.249} \approx 1.16 \mathrm{~m} $

So, $(S) \rightarrow(2)$

Therefore, the correct match is $\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 2$.

Thus, the correct option is (D).

To determine the amplitude and phase of the wave formed by the superposition of the two harmonic traveling waves, $ y_1(x, t) = 4 \sin (kx - \omega t) $ and $ y_2(x, t) = 2 \sin (kx - \omega t + \frac{2\pi}{3}) $, we use the following steps:

First, calculate the resultant amplitude $ A $ using the formula for the resultant of two waves with different phases:

$ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} $

where:

$ A_1 = 4 $ (amplitude of the first wave)

$ A_2 = 2 $ (amplitude of the second wave)

$ \phi = 120^\circ $ (phase difference, given by $\frac{2\pi}{3}$).

Substituting the values, we get:

$ A = \sqrt{4^2 + 2^2 + 2 \times 4 \times 2 \times \cos(120^\circ)} $

$ \cos(120^\circ) = -\frac{1}{2} $

Therefore:

$ A = \sqrt{16 + 4 + 16 \times (-\frac{1}{2})} $

$ A = \sqrt{16 + 4 - 8} $

$ A = \sqrt{12} = 2\sqrt{3} $

Next, compute the phase $ \phi $ of the resultant wave:

$ \tan \phi = \frac{A_2 \sin(\phi)}{A_1 + A_2 \cos(\phi)} $

$ \tan \phi = \frac{2 \sin(120^\circ)}{4 + 2 \cos(120^\circ)} $

$ \sin(120^\circ) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos(120^\circ) = -\frac{1}{2} $

Substituting these values, we get:

$ \tan \phi = \frac{2 \times \frac{\sqrt{3}}{2}}{4 + 2 \times -\frac{1}{2}} $

$ \tan \phi = \frac{\sqrt{3}}{3} $

This simplifies to:

$ \phi = \frac{\pi}{6} $

Thus, the amplitude and phase of the resultant wave are $ 2\sqrt{3} $ and $ \frac{\pi}{6} $ respectively.

To find the ratio of the velocities of transverse waves in two strings with different radii but identical materials and tension, consider the following:

The wave velocity $ v $ in a string is given by the formula:

$ v = \sqrt{\frac{T}{\mu}} $

where $ T $ is the tension and $ \mu $ is the linear mass density of the string, defined as:

$ \mu = \rho \pi R^2 $

Here, $ \rho $ is the density of the material, and $ R $ is the radius of the string. Given that both strings have the same tension $ T $ and material, we compare their wave velocities $ v_1 $ and $ v_2 $ for radii $ R_1 = R $ and $ R_2 = \frac{R}{2} $.

The velocity ratio is expressed as:

$ \frac{v_2}{v_1} = \frac{\sqrt{\frac{T}{\rho \pi R_2^2}}}{\sqrt{\frac{T}{\rho \pi R_1^2}}} $

Simplifying this expression:

$ \frac{v_2}{v_1} = \frac{\sqrt{R_1^2}}{\sqrt{R_2^2}} = \frac{R_1}{R_2} = \frac{R}{\frac{R}{2}} = 2 $

Thus, the ratio $ \frac{v_2}{v_1} $ is 2.

The equation for a wave traveling on a string is given by $ y = \sin(20\pi x + 10\pi t) $, where $ x $ is the distance and $ t $ is the time in SI units. To find the minimum distance between two points having the same oscillating speed, we use the concept that this distance is half the wavelength $\left(\frac{\lambda}{2}\right)$.

To find the wavelength $\lambda$:

$ \lambda = \frac{2\pi}{k} $

Here, the wave number $ k = 20\pi $. Thus,

$ \lambda = \frac{2\pi}{20\pi} = \frac{1}{10} \text{ m} = 10 \text{ cm} $

Thus, the minimum distance between two points having the same speed is:

$ \text{Distance} = \frac{\lambda}{2} = \frac{10 \text{ cm}}{2} = 5 \text{ cm} $

The superposition of the two harmonic waves results in the wave equation:

$ x = a \cos(1.5t) \cos(50.5t) $

This equation can be expanded using the trigonometric identity for the product of cosines:

$ x = \frac{a}{2} \cos[(1.5 + 50.5)t] + \frac{a}{2} \cos[(50.5 - 1.5)t] $

Simplifying, we find:

$ x = \frac{a}{2} \cos(52t) + \frac{a}{2} \cos(49t) $

This represents two waves with angular frequencies $52$ and $49$, respectively.

To find the beat frequency, we calculate the differences in frequencies:

$ f_1 = \frac{52}{2\pi}, \quad f_2 = \frac{49}{2\pi} $

The beat frequency is then:

$ f_{\text{Beat}} = f_1 - f_2 = \frac{3}{2\pi} \text{ Hz} $

The period of the beats, $T_{\text{Beat}}$, is the reciprocal of the beat frequency:

$ T_{\text{Beat}} = \frac{1}{f_{\text{Beat}}} = \frac{2\pi}{3} \text{ sec} $

Approximating this gives:

$ T_{\text{Beat}} \approx 2.09 \text{ sec} \approx 2 \text{ sec} $

$\begin{aligned} & \mathrm{x}(\mathrm{t})=5 \cos \left[628 \mathrm{t}+\frac{\pi}{2}\right] \mathrm{m} \\ & \text { velocity }\left(\mathrm{v}_\omega\right)=300 \mathrm{~m} / \mathrm{s} \\ & \mathrm{v}_{\mathrm{w}}=\frac{\omega}{\mathrm{K}} \\ & 300=\frac{628}{\mathrm{~K}} \Rightarrow \mathrm{~K}=\frac{628}{300} \\ & \frac{2 \pi}{\lambda}=\frac{628}{300} \Rightarrow \lambda=\frac{2 \times 3.14 \times 300}{628} \\ & \lambda=2 \mathrm{~m} \end{aligned}$

$\begin{aligned} &\lambda_1=4 \ell\\ &\mathrm{f}_1=\frac{\mathrm{v}}{4 \ell} \end{aligned}$

$\begin{aligned} & \lambda_2=\frac{16 \ell}{5} \\ & \mathrm{f}_2=\frac{5 \mathrm{~V}}{16 \ell} \\ & \frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{\frac{\mathrm{V}}{\ell}\left(\frac{1}{16}\right)}{\frac{\mathrm{V}}{4 \ell}} \times 100=25 \% \end{aligned}$

The fundamental frequency for a closed (organ) pipe can be expressed as:

$ f = \frac{v}{4\ell} $

For the first air column, with length $ \ell_1 $, the frequency $ f_1 $ is:

$ f_1 = \frac{v}{4\ell_1} $

For the second air column, with length $ \ell_2 $, the frequency $ f_2 $ is:

$ f_2 = \frac{v}{4\ell_2} $

The beat frequency, which is the difference in these two frequencies ($ f_1 - f_2 $), is given as 15 beats per second:

$ \text{Beat} = f_1 - f_2 = \frac{v}{4} \left( \frac{1}{\ell_1} - \frac{1}{\ell_2} \right) $

Substitute the given lengths into the formula:

$ 15 = \frac{v}{4} \left( \frac{1}{1} - \frac{1}{1.2} \right) $

Simplify the equation:

$ 15 = \frac{v}{4} \left( \frac{0.2}{1.2} \right) $

Solve for $ v $:

$ v = \frac{15 \times 4 \times 1.2}{0.2} = 60 \times 6 = 360 \, \text{m/s} $

Thus, the velocity of sound in the air column is 360 m/s.

To find the equation of the sinusoidal wave, we need to determine several properties of the wave:

Wave Velocity (v):

Given that the wave travels a distance of 1.2 cm in 0.3 seconds, the velocity $ v $ can be computed as:

$ v = \frac{\text{distance}}{\text{time}} = \frac{1.2 \, \text{cm}}{0.3 \, \text{s}} = 4 \, \text{cm/s} $

Wave Number (k):

The wave number $ k $ is calculated using the wavelength $ \lambda = 7.5 \, \text{cm} $:

$ k = \frac{2\pi}{\lambda} = \frac{2\pi}{7.5} = \frac{4\pi}{15} \approx 0.83 $

Angular Frequency ($\omega$):

Angular frequency $ \omega $ is related to the wave number and velocity by the equation $ v = \frac{\omega}{k} $, thus:

$ \omega = vk = 4 \times \frac{4\pi}{15} = \frac{16\pi}{15} \approx 3.35 $

Wave Equation:

The general form for a sinusoidal wave traveling in the positive x-direction is:

$ y = A \cos(kx - \omega t) $

Given the maximum displacement (amplitude $ A $) of the wave is 2 cm, the equation becomes:

$ y = 2 \cos(0.83x - 3.35t) \, \text{cm} $

This equation accurately describes the sinusoidal wave given the provided parameters.

We know, speed of sound in a medium $v = \sqrt {{B \over \rho }} $ where B = bulk modulus, $\rho$ = density of the medium

Since, solids and liquids are much more difficult to compress than gases so they have much higher values of bulk modulus.

i.e., ${B_{solid}} > {B_{liquid}} > {B_{gas}}$

Generally solids and liquids have higher mass densities ($\rho$) than gases. But corresponding increase in bulk modulus is much higher.

So, ${v_{solid}} > {v_{liquid}} > {v_{gas}}$

Hence, option C is correct.

Let's analyze the wave equation step by step.

The given wave is:

$y(x, t) = 4.0 \sin \left[20 \times 10^{-3} x + 600 t\right] \text{ mm}.$

First, simplify the coefficient of $ x $:

$20 \times 10^{-3} = 0.02 \, \text{mm}^{-1}.$

So the equation becomes:

$y(x, t) = 4.0 \sin\left(0.02x + 600t\right) \text{ mm}.$

A standard form for a travelling wave is:

$y(x, t) = A \sin(kx - \omega t)$

which represents a wave moving in the positive $ x $-direction with speed $ v = \frac{\omega}{k} $.

Notice that our wave equation has the form:

$\sin(0.02x + 600t)$

The positive sign in front of $ 600t $ means we can rewrite the phase as:

$0.02x + 600t = 0.02x - (-600t),$

which indicates that the angular frequency $ \omega $ in the standard form is effectively $ -600 $.

The velocity $ v $ of a wave is determined from the phase (for a constant phase, $ \phi = $ constant):

$k x + \omega t = \text{constant}.$

Differentiating with respect to $ t $:

$k \frac{dx}{dt} + \omega = 0,$

which gives:

$\frac{dx}{dt} = -\frac{\omega}{k}.$

Substituting the values:

$ k = 0.02 \, \text{mm}^{-1} $

$ \omega = 600 \, \text{s}^{-1} $

We have:

$v = -\frac{600}{0.02} = -30000 \text{ mm/s}.$

Convert the velocity from mm/s to m/s:

$-30000 \, \text{mm/s} = -30 \, \text{m/s}.$

Thus, the velocity of the wave is $-30 \, \text{m/s}$.

The correct answer is Option D.

We know, for closed pipe,

${f_n} = {{nv} \over {4l}},\,n = 1,3,5,7$

for open pipe, ${f_n} = {{nv} \over {2L}},\,n = 1,2,3,4$

So, 9$^{th}$ harmonic of closed pipe $ = {{9{v_1}} \over {4{l_1}}}$

4$^{th}$ harmonic of open pipe $ = {{4{v_2}} \over {2{l_2}}} = {{2{v_2}} \over {{l_2}}}$

$\therefore$ ${{9{v_1}} \over {4{l_1}}} = {{2{v_2}} \over {{l_2}}} \Rightarrow {{{l_2}} \over {{l_1}}} = {8 \over 9}{{{v_2}} \over {{v_1}}}$

We know, $v = \sqrt {{\beta \over \rho }} $

So, ${{{v_2}} \over {{v_1}}} = \sqrt {{{{\rho _1}} \over {{\rho _2}}}} $ (for same $\beta$)

hence, ${{{l_2}} \over {{l_1}}} = {8 \over 9}\sqrt {{{{\rho _1}} \over {{\rho _2}}}} = {8 \over 9}\sqrt {{1 \over {16}}} = {8 \over 9} \times {1 \over 4}$

$ \Rightarrow {l_2} = {2 \over 9} \times {l_1} \Rightarrow {l_2} = {2 \over 9} \times 10 = {{20} \over 9}cm$

$f_{\text {beat }}=\left|f_1-f_2\right|$

and $f_{\text {open }}=2 \times f_{\text {closed }}$

When tubes are vibrated with both of their ends open, then

$ \begin{aligned} & f_{\text {beat }}^{\prime}=2\left|f_1-f_2\right| \\ & f'_{\text {beat }}=2 \times 2 \\ & f_{\text {beat }}^{\prime}=4 \end{aligned} $

Frequency of echo heard by the driver is given by,

$ \begin{aligned} f^{\prime} & =\frac{v+u}{v-u} f \\ f^{\prime} & =\frac{v+u}{v-u} \times n \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Difference in frequencies is $0.1 n$

$ \begin{array}{ll} \therefore & f^{\prime}-n=0.1 n \\ & f^{\prime}=1.1 n \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i) and (ii)

$ \begin{aligned} & \frac{v+u}{v-u}=1.1 \\ & \frac{336+u}{336-u}=1.1 \\ & 2.1 u=33.6 \\ & u=16 \mathrm{~ms}^{-1} \end{aligned} $

The tube is 50 cm long and is closed at one end.

Finding the Wavelength:

For the fifth harmonic in a closed pipe, the formula is:

$ \frac{5 \lambda}{4} = L $

This means five-fourths of the wavelength fits into the total length ($L$) of the tube.

Let's solve for wavelength ($\lambda$):

$ \lambda = \frac{4L}{5} = \frac{4 \times 50}{5} = 40~\mathrm{cm} $

Calculating the Phase Difference:

The phase difference ($\Delta \phi$) between two points separated by a distance $x$ is:

$ \Delta \phi = \frac{2\pi}{\lambda} \times x $

Here, $x = 42 ~\mathrm{cm}$ (the distance from the open end to the other particle) and $\lambda = 40~\mathrm{cm}$:

$ \Delta \phi = \frac{2\pi}{40} \times 42 $

This can be calculated as:

$ \Delta \phi = \frac{2\pi \times 42}{40} = \frac{84\pi}{40} = \frac{21\pi}{10} $

If you convert this to degrees:

$ \Delta \phi = \frac{21\pi}{10} \times \frac{180^\circ}{\pi} = 378^\circ $

But phases repeat every 360°. So, we subtract 360° to get the simpler value:

$ 378^\circ - 360^\circ = 18^\circ $

So, the phase difference is $18^\circ$.

For fundamental frequency, at points $A$ and $B$, there should be an antinode and at point $C$, there should be a node.

$ \begin{aligned} \Rightarrow & & \frac{\lambda}{4}+\frac{\lambda}{4} & =L \\ \Rightarrow & & \lambda & =2 L=2 \times 125=250 \mathrm{~cm} \end{aligned} $

For frequency,

$ \begin{aligned} v & =\frac{v}{\lambda}=\frac{5000}{250 \times 10^{-2}} \\ & =2000 \mathrm{~Hz} \text { or } 2 \mathrm{kHz} \end{aligned} $

Beat frequency

$ f=|323-320|=3 \mathrm{~Hz} $

∴ Beat period

$ T=\frac{1}{f}=\frac{1}{3} \mathrm{~s} $

The time interval between a maximum sound and its adjacent minimum sound is equal to half of the beat period.

i.e., Time interval $=\frac{T}{2}=\frac{\frac{1}{3}}{2}=\frac{1}{6} \mathrm{~s}$

Step 1: Understanding the Doppler Effect for a Moving Observer

When the sound source is not moving and the observer moves toward it, the frequency they hear increases.

When the observer moves away from the stationary source, the frequency they hear decreases.

Step 2: Writing the Frequency Formulas

If the actual frequency of the source is $n$, the speed of sound is $v$, and the observer's speed is $u$:

When moving toward the source: $ n_1 = n \left( \frac{v + u}{v} \right) $

When moving away from the source: $ n_2 = n \left( \frac{v - u}{v} \right) $

Step 3: Using the Ratio Given in the Problem

We are told that the ratio of the two frequencies is $n_1 : n_2 = 71 : 65$. We set up an equation:

$ \frac{n_1}{n_2} = \frac{v + u}{v - u} = \frac{71}{65} $

Step 4: Solving for Observer's Speed $(u)$

Cross-multiply the equation: $ 65(v + u) = 71(v - u) $

Expand: $ 65v + 65u = 71v - 71u $

Move both $v$ terms to one side and $u$ terms to the other: $ 65u + 71u = 71v - 65v $ $ 136u = 6v $

Divide both sides by 136: $ u = \frac{6v}{136} = \frac{3v}{68} $

Step 5: Substitute Speed of Sound Value

The speed of sound is $v = 340~\mathrm{m/s}$.

Plug this value into the formula for $u$: $ u = \frac{3 \times 340}{68} = \frac{1020}{68} = 15~\mathrm{m/s} $

To change $15~\mathrm{m/s}$ into $\mathrm{km/h}$, multiply by $\frac{18}{5}$: $ 15 \times \frac{18}{5} = 54~\mathrm{km/h} $

Step 1: Find the amplitude.

The total distance from one extreme point to the other is 10 cm. So the amplitude is half of this: $5~\mathrm{cm} = 0.05~\mathrm{m}$.

Step 2: Find the angular frequency ($( \omega )$).

The frequency $f$ is $210~\mathrm{Hz}$.

Angular frequency: $\omega = 2\pi f = 2 \times \frac{22}{7} \times 210 = 44 \times 30 = 1320~\mathrm{rad/s}$.

Step 3: Find the wave number ($k$).

The speed $v$ is $330~\mathrm{m/s}$.

Wave number: $k = \frac{\omega}{v} = \frac{1320}{330} = 4~\mathrm{rad/m}$.

Step 4: Write the equation of the wave.

The standard equation for a wave traveling along the positive $X$-axis is $s(x, t) = A \sin(kx - \omega t)$.

Here, $A = 0.05~\mathrm{m}$, $k = 4~\mathrm{rad/m}$, and $\omega = 1320~\mathrm{rad/s}$, so:

$ \begin{aligned} s(x, t) &= 0.05 \sin(4x - 1320t) \end{aligned} $

Step 1: Formula for Fundamental Frequency

The basic formula for the frequency of a vibrating string is:

$ f = \frac{1}{2l} \sqrt{\frac{T}{\mu}} $

Here, $f$ is frequency, $l$ is the length of the string, $T$ is the tension, and $\mu$ is mass per unit length (which stays the same in this problem).

Step 2: Frequency with Initial Length and Tension

The first frequency, with length $l$ and tension $T_1$ is:

$ f_1 = \frac{1}{2l} \sqrt{\frac{T_1}{\mu}} \tag{i} $

Step 3: Frequency After Changing Length and Tension

The length of the string is shortened by $25\%$, so the new length is $75\%$ of $l$, or $\frac{3}{4}l$. The new tension is $T_2$ and the new frequency becomes $2f_1$ (it increases by $100\%$):

$ 2f_1 = \frac{1}{2(\frac{3}{4}l)} \sqrt{\frac{T_2}{\mu}} \tag{ii} $

Step 4: Set Up the Equation

Now, make it easier by expressing $2f_1$ in terms of the formula:

$ 2f_1 = \frac{1}{(3/2)l} \sqrt{\frac{T_2}{\mu}} $

Because splitting the denominator gives $2 \times \frac{1}{2(\frac{3}{4}l)} = \frac{1}{(3/2)l}$.

Step 5: Plug $f_1$ from Step 2 into the New Equation

Replace $f_1$ in the new formula using the old frequency formula:

$ 2 \left( \frac{1}{2l} \sqrt{\frac{T_1}{\mu}} \right) = \frac{1}{(3/2)l} \sqrt{\frac{T_2}{\mu}} $

Step 6: Simplify the Equation

Multiply both sides to clear out the denominators:

$ \frac{1}{l} \sqrt{\frac{T_1}{\mu}} = \frac{2}{3l} \sqrt{\frac{T_2}{\mu}} $

Multiply both sides by $3l$:

$ 3\sqrt{\frac{T_1}{\mu}} = 2\sqrt{\frac{T_2}{\mu}} $

Step 7: Get Rid of Square Roots by Squaring Both Sides

Square each side to eliminate the square roots:

$ 9\frac{T_1}{\mu} = 4\frac{T_2}{\mu} $

The $\mu$ cancels out:

$ 9T_1 = 4T_2 $

Step 8: Find $\frac{T_2}{T_1}$

Divide both sides by $4T_1$ to solve for $\frac{T_2}{T_1}$:

$ \frac{T_2}{T_1} = \frac{9}{4} $

Step 1: Write the formula for frequency in an open pipe

The frequency of the $n$th harmonic in an open organ pipe is:

$ f_0=\frac{n v}{2 l_0} $

Step 2: Find the frequency for the 3rd harmonic in the open pipe

For the third harmonic, $n = 3$:

$ f_0' = \frac{3 v}{2 l_0} $

Step 3: Write the formula for frequency in a closed pipe

The frequency of the $n$th harmonic in a closed pipe is:

$ f_c=\frac{n v}{4 l_c} $

Step 4: Find the frequency for the 5th harmonic in the closed pipe

For the fifth harmonic, $n = 5$:

$ f_c' = \frac{5 v}{4 l_c} $

Step 5: Write the ratio of the two frequencies

We want to find the ratio $\frac{f_0'}{f_c'}$.

$ \frac{f_0'}{f_c'} = \frac{\frac{3v}{2l_0}}{\frac{5v}{4l_c}} $

Step 6: Simplify the ratio

First, simplify by dividing:

$ \frac{3v}{2l_0} \times \frac{4l_c}{5v} = \frac{12l_c}{10l_0} $

Step 7: Use the ratio of lengths

It is given that the ratio of lengths of the open and closed pipes is $2:3$, so $\frac{l_0}{l_c} = \frac{2}{3}$.

Step 8: Substitute the value for $\frac{l_0}{l_c}$

$ \frac{12}{10} \times \frac{1}{\frac{l_0}{l_c}} = \frac{12}{10} \times \frac{1}{\frac{2}{3}} = \frac{12}{10} \times \frac{3}{2} = \frac{36}{20} = \frac{9}{5} $

Step 9: State the final ratio

The ratio of the frequencies is $9:5$.

The standard form of a travelling wave is

$ y=A \sin (k x \pm \omega t) $

Comparing

$ \begin{aligned} & y=3 \sin (4 x+200 t) \\ & k=4 \mathrm{rad} / \mathrm{m} \\ & \omega=\mathrm{rad} / \mathrm{sec} \end{aligned} $

wave speed is given by

$ v=\frac{\omega}{k}=\frac{200}{4}=50 \mathrm{~m} / \mathrm{s} $

Wave speed for string

$ \begin{aligned} & v=\sqrt{\frac{T}{\mu}} \Rightarrow \mu=\frac{T}{v^2} \\ & \mu=\frac{500}{(50)^2}=\frac{500}{2500}=0.2 \mathrm{~kg} / \mathrm{m} \end{aligned} $

The initial fundamental frequency

$ \begin{aligned} & f_1=\frac{1}{2 L_1} \sqrt{\frac{T_1}{\mu}} \\ & 300=\frac{1}{2 L_1} \sqrt{\frac{T_1}{\mu}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

The new fundamental frequency

$ \begin{aligned} & f_2=\frac{1}{2 L_2} \sqrt{\frac{T_2}{\mu}} \\ & 100=\frac{1}{2\left(2 L_1\right)} \sqrt{\frac{T_2}{\mu}} \\ & 100=\frac{1}{4 L_1} \sqrt{\frac{T_2}{\mu}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Divide the Eq. (i) by the Eq. (ii)

$ \begin{aligned} \frac{300}{100} & =\frac{\frac{1}{2 L_1} \sqrt{\frac{T_1}{\mu}}}{\frac{1}{4 L_1} \sqrt{\frac{T_2}{\mu}}} \\ 3 & =\frac{\frac{1}{2} \sqrt{T_1}}{\frac{1}{4} \sqrt{T_2}} \Rightarrow 3=\frac{4}{2} \sqrt{\frac{T_1}{T_2}} \\ 3 & =2 \sqrt{\frac{T_1}{T_2}}=\frac{3}{2}=\sqrt{\frac{T_1}{T_2}}=\frac{9}{4}=\frac{T_1}{T_2} \\ \frac{T_2}{T_1} & =\frac{4}{9} \end{aligned} $

The ratio of the new tension to the initial tension is $\frac{4}{9}$.

$I_R=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi$

Here, $\phi=\frac{\pi}{2}$ and $I_1=I_2=I$

$ \therefore \quad I_R=I+I+2 \sqrt{I \cdot I} \cos \frac{\pi}{2}=2 I $

Given:

Both source and observer are moving towards each other.

Each moves at 10% of the speed of sound.

$ v_s = 0.1v, \quad v_o = 0.1v $

where $ v $ = speed of sound.

Formula (for Doppler effect):

When both are moving towards each other, the apparent frequency $ f' $ heard by the observer is

$ f' = f \frac{v + v_o}{v - v_s} $

Substituting given values:

$ f' = f \frac{v + 0.1v}{v - 0.1v} = f \frac{1.1v}{0.9v} = f \times \frac{1.1}{0.9} $

$ \frac{1.1}{0.9} = 1.222... $

So,

$ f' = 1.222 f $

Percentage change in frequency:

$ \%\ \text{increase} = (1.222 - 1) \times 100 = 22.2\% $

✅ Final Answer:

$ \boxed{22.2\%} \text{ (Option C)} $

Velocity of sound wave

$ v=\frac{\text { Distance }}{\text { Time }}=\frac{600}{2}=300 \mathrm{~m} / \mathrm{s} $

∴ Wavelength, $\lambda=\frac{v}{f}=\frac{300}{500}=0.6 \mathrm{~m}$

Number of waves between the points $X$ and $Y=\frac{\text { Distance }}{\text { Wavelength }}=\frac{600}{0.6}=1000$

$ \begin{aligned} &\text { From the given wave equation }\\ &\begin{aligned} k & =5 \times 10^{-3} \mathrm{~cm}^{-1}=5 \times 10^{-1} \mathrm{~m}^{-1} \\ & =0.5 \mathrm{~m}^{-1} \\ w & =20 \mathrm{rad} / \mathrm{s} \\ \therefore \quad v & =\frac{\omega}{k}=\frac{20}{5 \times 10^{-1}}=40 \mathrm{~m} / \mathrm{s} \\ \therefore \quad \mu & =\frac{m}{l}=\frac{3 \times 10^{-3}}{0.8} \\ & =3.75 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \\ \therefore \text { Since, } v & =\sqrt{\frac{T}{\mu}} \\ \Rightarrow \quad v^2 & =\frac{T}{\mu} \\ \Rightarrow \quad T & =\mu v^2=3.75 \times 10^{-3} \times 40^2 \\ & =3.75 \times 10^{-3} \mathrm{~N}=6 \mathrm{~N} \end{aligned} \end{aligned} $

Standard equation of a travelling wave is,

$ y=A \sin (k x-\omega t) $

Given, $y=0.5 \sin (70.1 x-10 \pi t)$

$\therefore \omega=$ angular frequency $=10 \pi$

But $\omega=2 \pi f, f=$ frequency

$ \Rightarrow f=\frac{\omega}{2 \pi}=\frac{10 \pi}{2 \pi}=5 \mathrm{~Hz} $

Phase difference between waves is

$ \begin{aligned} \phi_1-\phi_2 & =\left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right)-\left(\omega t-\frac{2 \pi x}{\lambda}\right) \\ & =\phi \end{aligned} $

As, relation of phase and path difference is,

$ \frac{\Delta \phi}{2 \pi}=\frac{\Delta L}{\lambda} $

$\Rightarrow$ Path difference, $\Delta L=\frac{\lambda \phi}{2 \pi}$.

$ \begin{aligned} &\text { } y_1=3 \sin 250 \pi t\\ &\begin{array}{ll} \therefore & \omega_1=250 \pi \mathrm{rad} / \mathrm{s} \\ & y_2=2 \mathrm{sin} 260 \pi t \\ \therefore & \omega_2=260 \pi \mathrm{rad} / \mathrm{s} \\ \therefore & f_1=\frac{\omega_1}{2 \pi}=\frac{250 \pi}{2 \pi}=125 \mathrm{~Hz} \\ & f_2=\frac{\omega_2}{2 \pi}=\frac{260 \pi}{2 \pi}=130 \mathrm{~Hz} \\ & T=\frac{1}{f_2-f_1} \\ & =\frac{1}{130-125}=\frac{1}{5}=0.2 \mathrm{~s} \end{array} \end{aligned} $

In a closed organ pipe, the number of nodes formed in the $n$th harmonic is equal to $\left(\frac{n+1}{2}\right)$ if $n$ is odd.

Thus, for the fifth $(n=5)$ harmonic, the number of nodes $=\frac{5+1}{2}=3$

and for the ninth harmonic ( $n=9$ ), the number of nodes $=\frac{9+1}{2}=5$

Given:

A stretched wire has a fundamental frequency $ f $.

When it is divided into 3 segments, their fundamental frequencies are $ f_1, f_2, $ and $ f_3 $.

Tension $ T $ is constant.

We need to find the relation among $ f, f_1, f_2, f_3 $.

Step 1: Formula for fundamental frequency of a stretched string

For a string under tension $ T $, length $ L $, and linear density $ \mu $:

$ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} $

Step 2: Frequencies of the three parts

Suppose the original length is divided into three parts of lengths $ L_1, L_2, L_3 $, so that

$ L = L_1 + L_2 + L_3 $

Then their fundamental frequencies are

$ f_1 = \frac{1}{2L_1}\sqrt{\frac{T}{\mu}}, \quad f_2 = \frac{1}{2L_2}\sqrt{\frac{T}{\mu}}, \quad f_3 = \frac{1}{2L_3}\sqrt{\frac{T}{\mu}} $

Step 3: Express $ L_i $ in terms of $ f_i $

From the above,

$ L_i = \frac{1}{2f_i}\sqrt{\frac{T}{\mu}} $

and

$ L = \frac{1}{2f}\sqrt{\frac{T}{\mu}} $

Step 4: Use $ L = L_1 + L_2 + L_3 $

$ \frac{1}{2f}\sqrt{\frac{T}{\mu}} = \frac{1}{2}\sqrt{\frac{T}{\mu}} \left(\frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}\right) $

Cancel the common factor $ \frac{1}{2}\sqrt{\frac{T}{\mu}} $:

$ \boxed{\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}} $

✅ Correct Option: (C)

$ \boxed{\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}} $

Mass per unit length,

$ \mu=\frac{M}{l}=\frac{5 \times 10^{-3}}{0.81} $

$\therefore$ Speed of transverse wave

$ \begin{aligned} v & =\sqrt{\frac{T}{\mu}}=\sqrt{\frac{50}{\frac{5 \times 10^{-3}}{0.81}}} \\ & =\sqrt{\frac{50 \times 0.81}{5 \times 10^{-3}}}=90 \mathrm{~m} / \mathrm{s} \end{aligned} $

$y=0.7 \sin \left(\frac{7 \pi}{4} x\right) \cos (350 \pi t)$

Comparing with equation of stationary wave,

$y=2 A \sin (k x) \cos \omega t$

We get, $k=\frac{7 \pi}{4}, \omega=350 \pi$

$\therefore$ Speed of stationary wave,

$ v=\frac{\omega}{k}=\frac{350 \pi}{\frac{7 \pi}{4}}=200 \mathrm{~m} / \mathrm{s} $

$ \begin{aligned} \text { } f_1 & =\frac{v}{\lambda_1}=\frac{330}{0.99}=333.33 \mathrm{~Hz} \\ f_2 & =\frac{v}{\lambda_2}=\frac{330}{1}=330 \mathrm{~Hz} \\ \therefore f_{\text {beat }} & =f_1-f_2=333.33-330=3.33 \mathrm{~Hz} \\ \therefore \quad t & =\frac{\text { Total number of beat }}{f_{\text {beat }}}=\frac{10}{3.33}=3 \mathrm{~s} \end{aligned} $

To find the frequency of the plane progressive wave given by the equation $y = 2 \cos 2 \pi(330 \mathrm{t} - x) \mathrm{m}$, we start by analyzing the general form of a wave equation.

The general form of a wave equation is:

where:

• $A$ is the amplitude of the wave.
• $f$ is the frequency of the wave.
• $t$ is the time variable.
• $k$ is the wave number, defined as $\frac{2 \pi}{\lambda}$ where $\lambda$ is the wavelength.
• $x$ is the spatial variable.
• $\phi$ is the phase constant.

By comparing the given wave equation with the general form, we have:

We observe that the term $2 \pi(330t - x)$ corresponds to $2 \pi ft - kx$ in the general form.

From this, it is clear that:

• $2 \pi ft \rightarrow 2 \pi \cdot 330 t$
• Therefore, $f = 330$ Hz

So, the frequency of the wave is 330 Hz.

Thus, the correct answer is:

Option C: 330 Hz

$\begin{aligned} & \mathrm{f}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1} \\ & \mathrm{f}_1=\mathrm{f}_2 \\ & \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{\mathrm{v}}{\mathrm{L}_2} \\ & \Rightarrow \mathrm{L}_2=4 \mathrm{~L}_1 \\ & 60=4 \times \mathrm{L}_1 \\ & \mathrm{~L}_1=15 \mathrm{~cm} \end{aligned}$

Given,

Length of the wire, $ L=1 \mathrm{~m} $

Elastic strain, $ \varepsilon=1 \%=0.01 $

Density of metal, $ \rho=8000 \mathrm{~kg} / \mathrm{m}^{3} $

Young's modulus, $ Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m} $

We know that, $ Y=\frac{\sigma \text { (stress) }}{\varepsilon \text { (strain) }} $

$ \begin{array}{l} \sigma=Y \cdot \varepsilon=2 \times 10^{11} \times 0.01 \\\\ \sigma=2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} \end{array} $

We also new that,

$ \sigma=\frac{T}{A} \Rightarrow T=\sigma \times A $

Cross-sectional area is not given, so we consider mass per unit length $ \mu $ of the wire,

$ \mu=\rho \cdot A $

Putting in Eq. (i), we get

$ T=\frac{\sigma \cdot \mu}{\rho} $

The fundamental frequency of the transvèrse waves in the metal wire is,

$ \begin{array}{l} f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \\\\ f=\frac{1}{2 \times 1} \sqrt{\frac{\sigma \cdot \mu}{\rho \cdot \mu}}=\frac{1}{2} \sqrt{\frac{\sigma}{\rho}} \\\\ f=\frac{1}{2} \sqrt{\frac{2 \times 10^{9}}{8000}}=\frac{1}{2} \times \sqrt{\frac{10^{6}}{4}} \\\\ f=\frac{1}{4} \times 10^{3}=\frac{1000}{4} \\\\ f=250 \mathrm{~Hz} \end{array} $

Given,

Beat frequency, $ f_{\text {beat }}=6 \mathrm{~Hz} $

Length of the shorter pipe, $ L_{1}=150 \mathrm{~cm} $

$ =1.5 \mathrm{~m} $

Speed of sound in air, $ v=336 \mathrm{~m} / \mathrm{s} $

For a closed loop, fundamental frequency is given by,

$ f=\frac{v}{4 L} $

For the shorter pipe $ L_{1} $,

$ f_{1}=\frac{336}{4 \times 1.5}=\frac{336}{6}=56 \mathrm{~Hz} $

Let the length of longer pipe be $ L_{2} $ and frequency be $ f_{2} $.

Since the pipes produces 6 beats per second when sound together:

$ \left|f_{1}-f_{2}\right|=6 $

There are two possible cases for $ f_{2} $,

Case (i) $ f_{2}=f_{1}+6=56+6=62 \mathrm{~Hz} $

Case (ii) $ f_{2}=f_{1}-6=56-6=50 \mathrm{~Hz} $

So, from Eq. (i),

Case (i) $ f_{2}=\frac{v}{4 L_{2}} \Rightarrow 62=\frac{336}{4 L_{2}} $

$ \Rightarrow \quad L_{2}=\frac{336}{4 \times 62} $

$ L_{2}=1.35 \mathrm{~m}=135 \mathrm{~cm} $ (Not used because it is less than 4)

Case (ii) $ f_{2}=\frac{v}{4 L_{2}} \Rightarrow 50=L_{2}=\frac{336}{4 L_{2}} $

$ \Rightarrow \quad L_{2}=\frac{336}{4 \times 50} $

$ f_{2}=1.68 \mathrm{~m}=168 \mathrm{~cm} $

So, the length of the longer pipe that results in the production of 6 beats per second when sounded with the shorter pipe is 168 cm .

Given:

Path difference ($\Delta x$): 50 cm = 0.5 m

Phase difference ($\phi$): $1.8 \pi$

Velocity of sound ($v$): 340 m/s

We are to find the frequency of the sound wave.

First, use the relationship between phase difference, path difference, and wavelength:

$ \phi = \frac{2 \pi}{\lambda} \times \Delta x $

Substitute the given values:

$ 1.8 \pi = \frac{2 \pi}{\lambda} \times 0.5 $

Solve for $\lambda$ (wavelength):

$ 1.8 \pi = \frac{\pi}{\lambda} $

$ \lambda = \frac{1}{1.8} \, \text{m} $

Next, use the equation relating velocity, frequency, and wavelength:

$ v = f \times \lambda $

Substitute the known values:

$ 340 = f \times \frac{1}{1.8} $

Solve for $f$ (frequency):

$ f = 340 \times 1.8 $

$ f = 612 \, \text{Hz} $

Thus, the frequency of the sound wave is 612 Hz.

Given:

Frequency of source, $ f_{0} = 102 \, \text{Hz} $

Source velocity, $ v_{S} = 0 \, \text{m/s} $

Observers' Motion:

Each observer is moving with a velocity that is 10% of the speed of sound, i.e., $ v_{0} = 0.1v $.

Doppler Effect Calculation:

For each observer moving away from the source, the frequency heard is given by:

$ f_{\text{observer}} = f_{0} \left( \frac{v - v_{0}}{v} \right) $

where $ v $ is the speed of sound in the medium.

Frequency Calculation for Both Observers:

Since both observers are moving away from the source at the same speed and in opposite directions, the formula for frequency remains consistent for both:

$ f_{\text{observer 1}} = f_{0} \left( \frac{v - 0.1v}{v} \right) = f_{0} \left( \frac{0.9v}{v} \right) = 0.9 f_{0} $

$ f_{\text{observer 2}} = f_{0} \left( \frac{v - 0.1v}{v} \right) = f_{0} \left( \frac{0.9v}{v} \right) = 0.9 f_{0} $

Ratio of Frequencies:

The ratio of the frequencies heard by the two observers is:

$ \frac{f_{\text{observer 1}}}{f_{\text{observer 2}}} = \frac{0.9 f_{0}}{0.9 f_{0}} = 1:1 $

Thus, both observers hear the same frequency, resulting in a ratio of 1:1.

Fundamental frequency of open pipe

$ f_{O}=\frac{v}{2 L} $

Fundamental frequency of closed pipe

$ \begin{aligned} f_{C} & =\frac{v}{4 L} \\\\ \frac{v}{2 L}-\frac{v}{4 L} & =100 \mathrm{~Hz} \\\\ \frac{v}{4 L} & =100 \mathrm{~Hz} \end{aligned} $

(given)

Now, 2nd harmonic of open pipe

$ \begin{aligned} f & =\frac{2 v}{2 L} \end{aligned}\left(f_{n}=\frac{n v}{2 L}\right) $

3rd harmonic of closed pipe

$ (2 n-1) \frac{v}{4 L}=\frac{5 v}{4 L} $

Difference, $ \frac{5 v}{4 L}-\frac{v}{L}=\frac{v}{4 L} $

$ \frac{v}{4 L}=100 \mathrm{~Hz} $

$ y_{1}=5 \sin 400 \pi t, y_{2}=8 \sin 408 \pi t $

Compare with standard wave equation

$ \begin{aligned} y & =A \sin \omega t \\\\ \omega_{1} & =400 \pi, \omega_{2}=408 \pi \end{aligned} $

We know that, $ \omega=2 \pi f $

$ \begin{aligned} \omega_{1} & =2 \pi f_{1}=400 \pi \\\\ f_{1} & =200 \mathrm{~Hz} \\\\ \omega_{2} & =2 \pi f_{2}=408 \pi \\\\ f_{2} & =204 \mathrm{~Hz} \end{aligned} $

Number of beats/second

$ f_{2}-f_{1}=204-200=4 \mathrm{~Hz} $

Now, number of beats per minute

$ \begin{array}{l} =4 \times 60 \\\\ =240 \text { beats/minute } \end{array} $

Give velocity of train, $v_t=108 \mathrm{kmph}$

$ =108 \times \frac{5}{18}=30 \mathrm{~m} / \mathrm{s} $

Speed of sound in air, $v_s=330 \mathrm{~m} / \mathrm{s}$

From Doppler's effect,

$ f_0=\frac{v_s+v_{\text {obberver }}}{v_s+v_{\text {train }}} \times f_s $

When train is going way

$ \begin{aligned} \Rightarrow \quad f_1 & =\frac{330}{330+30} f_s \\\\ f_1 & =\frac{330}{360} f_s \\\\ f_1 & =0.9166 f_s \end{aligned} $

When train is closing in

$ \begin{aligned} f_2 & =\frac{330}{330-30} f_s \\\\ & =11 f_s \end{aligned} $

$\therefore \%$ change in frequency observed

$ \begin{aligned} & =\frac{11 f_s-0.9166 f_s}{11 f_s} \times 100 \% \\\\ & =0.1667 \times 100 \% \\\\ & =16.67 \% \end{aligned} $

Number of beats produced between $n$ and $n-1$

$ \begin{aligned} & =n-(n-1) \\\\ & =1 \end{aligned} $

Number of beats produced between $n$ and $n+1$

$ \begin{aligned} & =(n+1)-(n) \\\\ & =1 \end{aligned} $

Number of beats produced between $n+1$ and $n-1$

$ \begin{aligned} & =(n+1)-(n-1) \\\\ & =2 \end{aligned} $

Total number of beats produced

$ =1+1+2=4 $

Number of distinguishable beat per second,

$ (n+1)-(n-1)=2 $