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).

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$

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

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

Using the Doppler effect formula:

$f = f_0\left(\frac{c + v_0}{c + v_s}\right)$

where

• $f$ is the observed frequency (frequency heard by the passenger in car Q)
• $f_0$ is the source frequency (400 Hz)
• $c$ is the speed of sound (360 m/s)
• $v_0$ is the velocity of the observer (passenger in car Q) relative to the medium (air) in the direction of the source (40 m/s)
• $v_s$ is the velocity of the source (car P) relative to the medium (air) in the direction of the observer (20 m/s, since it's moving in the same direction as car Q)

Plugging in the values:

$f = 400\left(\frac{360 + 40}{360 + 20}\right)$

$f = 400\left(\frac{400}{380}\right)$

$f = 421$

So, the observed frequency is 421 Hz.

We can write the given equation as:

$y = \sqrt{(\sin\omega t)^2 + (\cos\omega t)^2} \cos\left(\omega t - \arctan\frac{\sin\omega t}{\cos\omega t}\right)$

Using the identity $\sin^2\theta + \cos^2\theta = 1$, we get:

$y = \sqrt{1 + \sin 2\omega t} \cos\left(\omega t - \frac{\pi}{4}\right)$

The amplitude of the motion is the maximum value of $|y|$, which occurs when $\sin 2\omega t = 1$, i.e., at $t = \frac{\pi}{4\omega} + \frac{n\pi}{\omega}$, where $n$ is an integer. Substituting this value of $t$ in the above equation, we get:

$|y_{\text{max}}| = \sqrt{1 + \sin \frac{\pi}{2}} = \sqrt{2}$

Therefore, the amplitude of the motion is $\sqrt{2}$

The phenomenon of frequency change due to relative motion between a source and an observer is called Doppler effect. However, if the observer is at rest relative to the source (as in this case where the passenger is inside the train), then the frequency heard by the observer is the same as the frequency produced by the source.

This is because the relative velocity between the source (whistle) and the observer (passenger in the train) is zero. The Doppler effect only applies when there is a relative velocity between the source and the observer.

Therefore, the frequency heard by the passenger inside the train is the same as the frequency of the whistle, i.e., 400 Hz.

So, the correct answer is:

400 Hz.

${v_{wave}} = \left| {{{coefficient\,of\,t} \over {coefficient\,of\,x}}} \right|$

$ = {{400} \over 1} = 400$ m/s

${\omega \over k} = A\omega $

$ \Rightarrow k = {1 \over A} = {1 \over {2\,cm}}$

$ \Rightarrow {{2\pi } \over \lambda } = {1 \over {2\,cm}}$

$ \Rightarrow \lambda = 4\pi \,cm$

We know, equation of wave travelling in positive x-direction is -

$y = A\sin (kx - wt)$

where $k = {{2\pi } \over \lambda }$

and $w = 2\pi f$

Here given $\lambda$ = 4 cm and frequency (f) = 100 Hz

$\therefore$ $k = {{2\pi } \over 4} = 0.5\pi $ cm^$-$1

and $w = 2\pi \times 100 = 200\pi $ s^$-$1

$\therefore$ Equation of travelling wave,

$y = A\sin (0.5\pi x - 200\pi t)$

Particle velocity = ${{\partial x} \over {\partial t}}$

$\Rightarrow$ Maximum particle velocity $ = (2\pi n)\,(10)$

$ \Rightarrow (2\pi n)\,(10) = (n\lambda )\,(4)$

$ \Rightarrow \lambda = 5\pi $

${v \over {4.08}} - {v \over {4.16}} = {{40} \over {12}}$

$v = {{40} \over {12}} \times {{4.08 \times 4.16} \over {0.08}}$

$ = 707.2$ m/s

Frequency is independent of medium. For denser medium, wavelength and speed both would decrease.

$f' = {f_0}\left[ {{{v - {v_0}} \over {v - {v_s}}}} \right]$

$ \Rightarrow f' = {f_0}\left[ {{{v + {v \over 5}} \over v}} \right]$

$ \Rightarrow f' = {{6{f_0}} \over 5}$

$\Rightarrow$ % change = 20

${y_1} = 5\sin (2\pi x - 2\pi vt)$

${y_2} = 3\sin (2\pi x - 2\pi vt + 3\pi )$

$\Rightarrow$ Phase difference = 3$\pi$

$ \Rightarrow {A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos (3\pi )} $

$ \Rightarrow {A_{net}} = 2$ cm