Waves

For closed organ pipe, fundamental frequency,

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

For open organ pipe, first overtone frequency,

$ f_2=\frac{2 v}{2 l_0}=\frac{v}{l_0} $

$ \begin{aligned} &\text { Now, } f_1=f_2 \Rightarrow \frac{v}{4 l_c}=\frac{v}{l_0}\\ &\therefore l_c=\frac{l_0}{4}=\frac{50}{4}=12.5 \mathrm{~cm} \end{aligned} $

The distance of a oscillating point from a node is

$ y=A \sin \left(\frac{20}{3} \pi x\right) $

$ \begin{aligned} & \text { For } y=\frac{A}{2}, \sin \left(\frac{20}{3} \pi x\right)=\frac{1}{2} \\ & \Rightarrow \quad \frac{20 \pi x}{3}=\frac{\pi}{6} \\ & \therefore \quad x=\frac{1}{40} \mathrm{~m}=\frac{100}{40} \mathrm{~cm}=2.5 \mathrm{~cm} \end{aligned} $

∴ Distance between two successive oscillating points is

$ D=2 x=2 \times 2.5=5 \mathrm{~cm} $

To solve this problem, we need to understand the relationship between the frequencies of open and closed organ pipes and how their lengths are connected to these frequencies.

Firstly, let’s consider the open organ pipe. The fundamental frequency of an open organ pipe is given by:

$ f_1 = \frac{v}{2L_1} $

where $ f_1 = 600 \, \text{Hz} $ is the fundamental frequency, $ v = 330 \, \text{m/s} $ is the speed of sound, and $ L_1 $ is the length of the open organ pipe.

Hence:

$ 600 = \frac{330}{2L_1} $

Solving for $ L_1 $, we get:

$ L_1 = \frac{330}{2 \times 600} = \frac{330}{1200} = 0.275 \, \text{m} = 27.5 \, \text{cm} $

For the open organ pipe, the first overtone frequency will be:

$ f_2 = \frac{2v}{2L_1} = \frac{v}{L_1} = \frac{330}{0.275} \, \text{m} = 1200 \, \text{Hz} $

Now, considering the closed organ pipe, the first overtone frequency for the closed organ pipe (which corresponds to the third harmonic) is given by:

$ f'_2 = \frac{3v}{4L_2} $

We are given that the first overtone of the open organ pipe (which is $ 1200 \, \text{Hz} $) is equal to the first overtone of the closed organ pipe. Hence:

$ 1200 = \frac{3 \times 330}{4L_2} $

Solving for $ L_2 $, we get:

$ 1200 = \frac{990}{4L_2} $

Rearranging:

$ 4L_2 = \frac{990}{1200} $

$ L_2 = \frac{990}{4800} \, \text{m} = 0.20625 \, \text{m} = 20.625 \, \text{cm} $

Given the options, there appears to be a minor discrepancy. However, the closest value is:

Option A: 21 cm

Beats will be formed as the wavelengths (hence frequencies) are similar.

$\begin{aligned} & \lambda_1=40 \mathrm{~cm} ~\lambda_2=42 \mathrm{~cm} \\ & v \propto \sqrt{T} \\ & \Rightarrow \quad \frac{v_1}{v_2}=\left(\frac{T_2}{T_1}\right)^{1 / 2}=\left(\frac{273+20}{273}\right) \\ & \Rightarrow \quad v_2=319.23 \\ & v_1=\frac{v_1}{\lambda_1}=\frac{319.23}{0.40}=798.075 \mathrm{~Hz} \\ & v_2=\frac{v_2}{\lambda_2}=\frac{319.23}{0.42}=760.071 \mathrm{~Hz} . \end{aligned}$

No. of beats $=v_2-v_1=798-760=38 \mathrm{~Hz}$

The fundamental frequency of a string is inversely proportional to its length. So, if the string is divided into four segments of lengths $l_1, l_2, l_3$ and $l_4$, then the fundamental frequencies of these segments will be inversely proportional to their lengths.

Let the original length of the string be $L$. Then, we have:

$l_1 + l_2 + l_3 + l_4 = L$

The fundamental frequency of the original string is given by:

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

where $T$ is the tension in the string and $\mu$ is the mass per unit length of the string.

The fundamental frequencies of the four segments are given by:

$v_1 = \frac{1}{2l_1} \sqrt{\frac{T}{\mu}}$

$v_2 = \frac{1}{2l_2} \sqrt{\frac{T}{\mu}}$

$v_3 = \frac{1}{2l_3} \sqrt{\frac{T}{\mu}}$

$v_4 = \frac{1}{2l_4} \sqrt{\frac{T}{\mu}}$

Adding these equations, we get:

$v_1 + v_2 + v_3 + v_4 = \frac{1}{2} \sqrt{\frac{T}{\mu}} \left(\frac{1}{l_1} + \frac{1}{l_2} + \frac{1}{l_3} + \frac{1}{l_4} \right)$

Now, using the equation $l_1 + l_2 + l_3 + l_4 = L$, we can write:

$\frac{1}{l_1} + \frac{1}{l_2} + \frac{1}{l_3} + \frac{1}{l_4} = \frac{L}{l_1 l_2 l_3 l_4}$

Substituting this into the previous equation, we get:

$v_1 + v_2 + v_3 + v_4 = \frac{1}{2} \sqrt{\frac{T}{\mu}} \frac{L}{l_1 l_2 l_3 l_4}$

Since $v = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, we can write:

$v_1 + v_2 + v_3 + v_4 = \frac{vL}{l_1 l_2 l_3 l_4}$

Therefore, the correct answer is:

$\boxed{\frac{1}{v} = \frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3} + \frac{1}{v_4}}$

This is option C.

Given, first resonance column at depth, ${l_1} = 24$ cm

Second resonance column at depth, ${l_2} = 78$ cm

Third resonance column at depth, ${l_3} = ?$

As

${l_2} + x = {{3\lambda } \over 4} = 78$ ........ (ii)

${l_3} + x = {{5\lambda } \over 4}$ ........ (iii)

From Eqs. (i) and (ii),

$x = {{{l_2} - 3{l_1}} \over 2} = {{78 - 3(24)} \over 2} = 3$ ....... (iv)

From Eqs. (iii) and (i), ${{{l_3} + x} \over {{l_1} + x}} = 5$

${l_3} = 5{l_1} + 5x - x = 5{l_1} + 4x$

Substituting $l_1$ and x value, we get

${l_3} = 5(24) + 4(3) \Rightarrow {l_3} = 132$ cm.

Given, velocity of submarine A, v_A = 17 m/s

Velocity of submarine B, v_B = 34 m/s

Signal sent by submarine B is detected by submarine A can be shown as

Frequency of signal, f₀ = 600 Hz

Speed of sound in water v_s = 1500 ms^$-$1

Frequency received by submarine A is

${f_1} = \left( {{{{v_s} - {v_A}} \over {{v_s} - {v_B}}}} \right){f_0} = \left[ {{{1500 - 17} \over {1500 - 34}}} \right] \times 600$ Hz

${f_1} = {{1483} \over {1466}} \times 600$ ..... (i)

Frequency received by submarine B is

${f_2} = \left( {{{{v_s} + {v_B}} \over {{v_s} + {v_A}}}} \right){f_1}$

Substituting given values and f₁ value from Eq. (i), we get

${f_2} = \left( {{{1500 + 34} \over {1500 + 17}}} \right) \times \left[ {{{1483} \over {1466}} \times 600} \right]$

${f_2} = 1.0112 \times 1.0115 \times 600$

${f_2} = 613.7$ Hz.

The velocity of transverse waves is given by $v = \sqrt {{T \over m}} $ where, T = Tension and m = mass per unit length of the wire. If r is the radius of the wire and $\rho$ its density then,

$m = \pi {r^2}\rho $

$\therefore$ $v = \sqrt {{T \over m}} = \sqrt {{T \over {\pi {r^2}\rho }}} $

$\therefore$ ${v_A} = {{\sqrt {{T_A}} } \over {{r_A}\sqrt {\pi \rho } }}$ and

${v_B} = {{\sqrt {{T_B}} } \over {{r_B}\sqrt {\pi \rho } }}$

Now, ${{{v_A}} \over {{v_B}}} = \sqrt {{{{T_A}} \over {{T_B}}}} .\,{{{r_B}} \over {{r_A}}}$

$\because$ ${r_A} = 2{r_B}$ and ${T_A} = {1 \over 2}{T_B}$

Hence, ${{{v_A}} \over {{v_B}}} = {1 \over {2\sqrt 2 }}$

Let the speed of wave in string be v and tension in string be T.

$v \propto \sqrt T $

${{{v_{AB}}} \over {{v_{CD}}}} = \sqrt {{T \over {2T}}} = {1 \over {\sqrt 2 }}$

Given,

$Y = \sin \left[ {x\left( {{t \over 5} - {x \over 9}} \right) + {\pi \over 6}} \right]$ ...... (i)

The general equation of progressive wave is

$Y = A\sin \left[ {2\pi \left( {{t \over T} - {x \over \lambda }} \right) + \phi } \right]$ ...... (ii)

On comparing Eq. (i) with Eq. (ii), we get

A = 1 cm, f = 0.1 Hz, v = 1.8 cm/s and ${{2\pi } \over \lambda } = {\pi \over 9}$

$\Rightarrow$ $\lambda$ = 18 cm

Apparent frequency of car A = Apparent frequency of car B

$ \Rightarrow {v \over {v - {v_s}}} \times \nu = {v \over {v - v{'_s}}} \times \nu '$

$ \Rightarrow {\nu \over {v - 15}} = {{504} \over {v - 30}}$

$ \Rightarrow \nu = {{340 - 15} \over {340 - 30}} \times 504$ [$\because$ velocity of sound in air is 340 m/s]

$ \Rightarrow \nu = 529.2$ Hz

The given equation is

$y = 4\sin \left( {{{\pi x} \over {15}}} \right)\cos (96\pi t)$ ...... (i)

This equation is similar to the displacement equation of a travelling wave, $y = A\sin (\omega t - kx)$

On comparing these two equations, we get

$\omega = 2\pi v = 96\pi $

or, $\nu = {{96\pi } \over {2\pi }} = 48$ Hz

Now, the velocity of the string at a point x at time t is obtained by differentiating Eq. (i) w.r.t.

$\therefore$ $v = {{dy} \over {dt}} = - (4 \times 9\pi ) \times \sin \left( {{{\pi x} \over {15}}} \right)\sin (96\pi t)$

At x = 7.5 cm = 7.5 $\times$ 10^$-$2 m and t = 0.25 s,

$v = - 384\pi \sin \left( {\pi \times {{7.5 \times {{10}^{ - 2}}} \over {15}}} \right)\sin \left( {96\pi \times {{25} \over {100}}} \right)$

$ = - 384\pi \sin \left( {{\pi \over {200}}} \right)\sin (24\pi )$

$ = 0$ ($\because$ $\sin 24\pi = 0$)

Hence, the velocity of the particle is zero.