Waves

Transverse waves are those waves whose vibration is perpendicular to direction of propagation.

It can propagate through solid and liquid. Longitudinal waves are those wave's whose vibration of medium is parallel to direction of propagation.

It can propagate in all mediums (solid, liquid or gases).

No mechanical waves can propagate without any medium.

So, option (c) is the correct option.

Given,

Observer velocity, $v_O=10 \mathrm{~m} / \mathrm{s}$

Source velocity, $v_S=-10 \mathrm{~m} / \mathrm{s}$

Speed of sound $v=340 \mathrm{~m} / \mathrm{s}$

Source frequency $f^{\prime}=1980 \mathrm{~Hz}$

According to Doppler's effect formula

$ \begin{aligned} f^{\prime} & =f\left(\frac{v-v_0}{v-v_s}\right) \\ 1980 & =f\left(\frac{340-10}{340+10}\right) \\ f & =\frac{1980 \times 350}{330} \\ f & =2100 \mathrm{~Hz} \end{aligned} $

Given, displacement

$ y=5 \times 10^3 \sin \left(125 \pi x-\frac{\pi}{2} t\right) $

comparing it with standard wave equation,

$ \begin{aligned} & y=A \sin (\omega t+k x), \text { we get } \\ & A=5 \times 10^{-3} \mathrm{~m} \\ & K=125 \pi \\ & \omega=\frac{\pi}{2} \end{aligned} $

We know that, $K=\frac{2 \pi}{\lambda}$

Thus, $\lambda=\frac{2 \pi}{K}=\frac{2 \pi}{125 \pi}=0.16 \mathrm{~m}$

Also $T=\frac{2 \pi}{\omega}$

Thus, $T=\frac{2 \pi}{\pi / 2}=4 \mathrm{~s}$

Given, $v_A=250 \mathrm{~Hz}$

Beats $=5$ beats $/ \mathrm{s}$

$ \begin{array}{ll} & v_2-v_1=5 \\ \Rightarrow & v_2=5+250=255 \mathrm{~Hz} \end{array} $

For closed organ pipe $v_1=\frac{v}{4 L_1}$

For 7 th harmonic; $7 v_1=\frac{7 v}{4 L_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For an open pipe, fundamental energy is $v=\frac{v}{2 L_2}$

For 7th harmonic of open pipe

$ 4 v_2=\frac{4 v}{2 L_2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} &\text { Therefore from Eqs. (i), and (ii) we have }\\ &\begin{aligned} 7 v_1 & =4 v_2 \\ \Rightarrow \quad \frac{7 v}{4 L_1} & =\frac{4 v}{2 L_2} \\ & =\frac{L_1}{L_2}=\frac{14}{16}=\frac{7}{8}=7: 8 \end{aligned} \end{aligned} $

Apparent frequency is given by

$ f=f_0\left(\frac{v_s+v_0}{v_S}\right)=f_0\left(\frac{v+\frac{v}{S}}{5}\right)=\frac{6}{5} f_0 $

Hence, percentage increase in frequency

$ =\frac{\left(\frac{6}{5} f_0-f_0\right)}{f_0} \times 100 \%=20 \% $

Given,

Length of $\operatorname{rod} B D=L$

Length of string $A B=C D=1$

Mass suspended $=M$

Length of $B O=x$

$ \text { Length of } O D=L-x $

Frequency of 1st harmonics of $A B=$ Frequency of 2nd harmonics of $C D$. Let $T_{A B}$ is tension in string $A B . T_{C D}$ is the tension in string $C D$.

Frequency of 1st harmonics of $A B$ is given by

$ f=\frac{1}{2 l} \sqrt{\frac{T_{A B}}{m}}, l=\text { length of string }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Frequency of 2nd harmonics of $C D$

$ f=\frac{2}{2} \sqrt{\frac{T_{C D}}{m}}=\frac{1}{l} \sqrt{\frac{T_{C D}}{m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

Eq. (i) $=$ Eq. (ii) (given)

$ \begin{aligned} \Rightarrow & \frac{1}{2 l} \sqrt{\frac{T_{A B}}{m}} =\frac{1}{l} \sqrt{\frac{T_{C D}}{m}} \\ \Rightarrow & T_{A B} =4 T_{C D}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

For rotational equilibrium of massless rod, taking torque at point $O$.

$ \begin{aligned} & \tau=r \times F(\text { Here } F=\text { Tension }) \\ & T_{A B} \times(X)=T_{C D}(L-X) \end{aligned} $

Put value of $T_{A B}$, then

$ X=L / 5 $

When an observer moves towards a stationary sound source, the apparent frequency they perceive increases. This is due to the observer's motion, which causes them to encounter sound waves more frequently than if they were stationary.

In this scenario, the observer's speed is $\frac{1}{5}$ of the speed of sound. The key parameters are:

Observer's speed, $v_0 = \frac{v}{5}$

Source speed, $v_s = 0$ (stationary)

Frequency of the sound emitted by the source, $f$

Speed of sound, $v$

The formula to calculate the apparent frequency ($f'$) when the observer is moving towards a stationary source is:

$ f' = \left(\frac{v + v_0}{v}\right) f $

Substituting the given values:

$ f' = \left(\frac{v + \frac{v}{5}}{v}\right) f $

Simplifying the equation:

$ f' = \left(\frac{6v}{5v}\right) f = \frac{6}{5}f = 1.2f $

Therefore, the apparent frequency recorded by the observer is $1.2f$.

Speed of wave on the string is given by

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

where $T$ is tension, $\mu$ is linear mass density.Mass of the rope $=m$

$ \mu=\frac{m}{L} $

, iii)

Put the value of $T$ and $\mu$ in Eq. (i),

$ v=\sqrt{\frac{m g}{\frac{m}{L}}}=\sqrt{g L} $

Now, using equation of motion,

$ v^2=u^2+2 a s $

( $v=0$, rope is connected to ceiling)

$ \begin{aligned} & a=\frac{U^2}{2 S} \Rightarrow \frac{g L}{2 L} \\ & \quad \quad \quad \text { (because } S=L \text { ) } \\ & a=\frac{g}{2} \end{aligned} $

Hence, the value of acceleration is constant.

Let's analyze the given wave equation. The equation is:

$ y = (0.02) \sin (\pi x - 8\pi t) $

Here, the wave is in the standard form:

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

where:

$A = 0.02$ is the amplitude.

$k$ is the wave number.

$\omega$ is the angular frequency.

From the given equation:

The wave number is: $k = \pi$.

The angular frequency is: $\omega = 8\pi$.

The wave velocity $v$ (also called the phase velocity) is given by the formula:

$ v = \frac{\omega}{k} $

Substitute the known values:

$ v = \frac{8\pi}{\pi} = 8 \, \text{ms}^{-1} $

Thus, the velocity of the wave is $8 \, \text{ms}^{-1}$, which corresponds to Option C.

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.

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

$ = {{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

We know that, fundamental frequency produced in cylindrical open tube

$ f=\frac{V}{2 L} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where $L=$ length of tube When tube is dipped vertically in water so that half of its portion is in water, then it acts a closed pipe of length half of original length $\left(\frac{L}{2}\right)$.

So, fundamental frequency of this closed tube,

$ f^{\prime}=\frac{v}{4\left(\frac{L}{2}\right)}=\frac{v}{2 L}=f \,\,\,\,\,[from Eq. (i)]$

Maximum speed of wave is given as

$ \begin{aligned} v_{\max } & =\omega A \\ & =\text { Angular speed } \text { × } \text { Amplitude } \end{aligned} $

Thus,

In option (a) $y(x, t)=2 \sin (2 x-2 t)$

$ \begin{aligned} & & A & =2, \omega=2 \\ & \therefore & v_{\text {max }} & =2 \times 2=4 \mathrm{~m} / \mathrm{s} \end{aligned} $

In option (b), $y(x, t)=3 \sin (2 x-3 t)$

$ \begin{array}{rlrl} & & A & =3, \omega=3 \\ \therefore & & v_{\max } & =3 \times 3 =9 \mathrm{~m} / \mathrm{s} \end{array} $

In option (c), $y(x, t)=2 \sin (3 x-2 t)$

$ \begin{aligned} A & =2, \omega=2 \\ \therefore \quad v_{\max } & =2 \times 2=4 \mathrm{~m} / \mathrm{s} \end{aligned} $

In option (d), $y(x, t)=3 \sin (5 x-2 t)$

$ \begin{aligned} A & =3, \omega=2 \\ \therefore \quad v_{\max } & =3 \times 2=6 \mathrm{~m} / \mathrm{s} \end{aligned} $

Clearly, wave speed is mximum as 9 $\mathrm{m} / \mathrm{s}$ in wave equation given in option (b).

Initial length of wire, $I_1=0.4 \mathrm{~m}$

Frequency of stretched wire,

$ f_1=250 \mathrm{vib} / \mathrm{s}=250 \mathrm{~Hz} $

Final length of wire after stretching.

$ \begin{aligned} & l_2=l_1+0.1 \mathrm{~m} \\ & =0.4+0.1=0.5 \mathrm{~m} \end{aligned} $

Initial force (tension), $T_1=T$

Final force (tension), $T_2=\frac{T}{4}$

We know that, frequency of wave produced due to stretched string.

$ \begin{aligned} f & =\frac{l}{2} \sqrt{\frac{T}{m}} \\ \Rightarrow \quad \frac{f_2}{f_1} & =\frac{l_1}{l_2} \cdot \sqrt{\frac{T_2}{T_1}}=\frac{0.4}{0.5} \sqrt{\frac{T / 4}{T}} \\ & =\frac{4}{5} \times \frac{1}{2} \\ \frac{f_2}{f_1} & =\frac{2}{5} \\ \Rightarrow \quad f_2 & =\frac{2}{5} f_1=\frac{2}{5} \times 250=100 \mathrm{~Hz} \end{aligned} $

Beat frequency produced by two strings $A$ and $B=\Delta f_1$

When tension in string $A$ is slightly increased, then beat of frequency $=\Delta f_2$ Since, on increasing tension in the string $A$, beat frequency $\Delta f_2$ decreases from initial beat frequency $\Delta f_1$. Hence, frequency of string $B$ must be greater than fundamental frequency of string A. i.e, frequency of string $B=f_0+\Delta f_1$

The distance between two successive minima is equal to wavelength of transverse wave.

l.e. $\lambda=2.7 \mathrm{~m}$

Distance travelled by wave in 15 s is equal to $5 \lambda$ as five crests of the wave pass a given point.

i.e $d=\lambda=5 \times 2.7=13.5 \mathrm{~m}$

and $t=15 \mathrm{~s}$

∴ Speed of wave, $v=\frac{d}{t}=\frac{13.5}{1.5}=0.9 \mathrm{~m} / \mathrm{s}$

The ratio between the maximum and minimum intensities of the resultant wave,

$ \begin{aligned} & I_{\max }: I_{\min }=9: 4 \\ & \Rightarrow \frac{I_{\max }}{I_{\min }}=\frac{9}{4} \Rightarrow \frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2}=\left(\frac{3}{2}\right)^2 \\ & \Rightarrow \frac{A_1^2\left(1+\frac{A_2}{A_1}\right)^2}{A_1^2\left(1-\frac{A_2}{A_1}\right)^2}=\left(\frac{3}{2}\right)^2 \\ & \Rightarrow \frac{\left(1+\frac{A_2}{A_1}\right)^2}{\left(1-\frac{A_2}{A_1}\right)^2}=\left(\frac{3}{2}\right)^2 \Rightarrow \frac{1+\frac{A_2}{A_1}}{1-\frac{A_2}{A_1}}=\frac{3}{2} \\ & \Rightarrow \frac{A_1+A_2}{A_1-A_2}=\frac{3}{2} \\ & \Rightarrow 2 A_1+2 A_2=3 A_1-3 A_2 \\ & \Rightarrow 5 A_2=A_1 \Rightarrow \frac{A_2}{A_1}=\frac{1}{5}=0.20 \end{aligned} $

Here, both cars are moving towards each other, hence one car is like an observer and another car is like a source.

$\therefore \quad v_s=50 \mathrm{~m} / \mathrm{s} \Rightarrow v_0=-50 \mathrm{~m} / \mathrm{s}$

Frequency of horn blown by the source car,

$v=250 \mathrm{~Hz}$

According to Doppler's effect, frequency heared by another car $v^{\prime}$ is given as

$v^{\prime}=\frac{v-v_0}{v-v_s} \times v=\frac{350-(-50)}{350-50} \times 250$

$\therefore(v=$ speed of sound $=350 \mathrm{~m} / \mathrm{s})$

$=\frac{400 \times 250}{300} \Rightarrow v^{\prime}=\frac{100}{3} \mathrm{~Hz}$

Wavelength of heard sound,

$\begin{aligned} \lambda^{\prime} & =\frac{v}{v^{\prime}} \\ & =\frac{350}{1000 / 3}=\frac{1050}{1000}=1.05 \mathrm{~m}=105 \mathrm{~cm} \end{aligned}$

The speed of sound in air at room temperature is approximately $ 340 \, \text{m/s} $ or $ 3.4 \times 10^2 \, \text{m/s} $.

Given, mass of string, $m=2 \mathrm{~g}$

$=2 \times 10^{-3} \mathrm{~kg}$

length of the string, $l=1 \mathrm{~m}$

$\therefore$ Mass per unit length

$\mu=\frac{m}{l}=2 \times 10^{-3} \mathrm{~kg} / \mathrm{m}$

Tension in the string,

$\begin{aligned} & T=\text { mass of body } \times g \\ & T=m^{\prime} g \end{aligned}$

Frequency of transverse wave is given as

$\begin{aligned} f & =\frac{1}{21} \sqrt{\frac{T}{\mu}} \\ \Rightarrow \quad 100 & =\frac{1}{2 \times 1} \sqrt{\frac{m^{\prime} g}{2 \times 10^{-3}}} \quad(\because \text { Given, } f=100 \mathrm{~Hz}) \\ \Rightarrow(200)^2 & =\frac{m^{\prime} g}{2 \times 10^{-3}} \\ \Rightarrow \quad m^{\prime} & =\frac{4 \times 10^4 \times 2 \times 10^{-3}}{g}=\frac{80}{10}=8 \mathrm{~kg} \end{aligned}$

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 $

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