Waves

y₁ = 10^–6sin[100t + (x/50) + 0.5]

y₂ = 10^–6cos[100t + (x/50)]
[using cosx = sin(x + $\pi $/2)]

= 10^–6sin[100t + (x/50) + $\pi $/2]

= 10^–6sin[100t + (x/50) + 1.57]

The phase difference = 1.57 – 0.5 = 1.07

[or using sinx = cos($\pi $/2 – x). We get the same result.]

Apparent frequency, $f' = {{v + {v_0}} \over v}f$

$ = {{v + (1/5)v} \over v}f = 1.2f$

Wavelength does not change by motion of observer.

The whistle is revolving in a circle of radius 50 cm. So the source (whistle) is moving and the observer is fixed.


The minimum frequency will be heard by the observer when the linear velocity of the whistle (source) will be in a direction as shown in the figure, i.e. when the source is receding.

The apparent frequency heard by the observer is then given by $\upsilon ' = \upsilon \left( {{v \over {V + v}}} \right)$

where V and v are the velocities of sound and source respectively and $\upsilon $ is the actual frequency.

Now $v = r\omega = 0.5 \times 20 = 10\,m/s$

V = 340 m/s, $\upsilon $ = 385 Hz.

$ \therefore $ $\upsilon ' = 385 \times {{340} \over {340 + 10}} = 374$ Hz

The equation of progressive wave travelling in positive x-direction is given by

$y = a\sin {{2\pi } \over \lambda }\left( {vt - x} \right)$

Here a = 0.2 m, v = 360 m/sec, $\lambda $ = 60 m

$ \therefore y = 0.2\sin {{2\pi } \over {60}}\left( {360t - x} \right)$

$ = 0.2sin\left[ {2\pi \left( {6t - {x \over {60}}} \right)} \right]$

Comparing the given equation with general equation,
$y = a\sin 2\pi \left( {{t \over T} - {x \over \lambda }} \right)$, we get

$T = {{2\pi } \over {100}}$ and $\lambda = 20\pi $

$ \therefore v = \upsilon \lambda = {{100} \over {2\pi }} \times 20\pi = 1000\,m/s$

Resultant amplitude = 2a (1 + cos$\phi $) = a

$ \therefore $ $\left( {1 + \cos \phi } \right) = 1/2;$

$cos\phi = - {1 \over 2};\phi = {{2\pi } \over 3}$

$n = {1 \over {2l}}\sqrt {{T \over {\pi {r^2}\rho }}} $

$\rho _1^{'} = {\rho \over 2};T' = 2T$

D' = 2D or r' = 2r

$n' = {1 \over {2l}}\sqrt {{{2T} \over {\pi {{\left( {2r} \right)}^2}{\rho \over 2}}}} $

After solving, $n' = {1 \over {2l}}\sqrt {{{2T} \over {\pi {r^2}l}}} = n$

As $n \propto \left( {1/l} \right)$

$l = {l_1} + {l_2} + {l_3}$

$ \therefore {1 \over n} = {1 \over {{n_1}}} + {1 \over {{n_2}}} + {1 \over {{n_3}}}$

Given : x = acos($\omega $t + $\delta $)
and y = acos($\omega $t + $\alpha $)     ...(i)

where, $\delta = \alpha + \pi /2$

$ \therefore {\rm{x = acos(}}\omega {\rm{t + }}\alpha {\rm{ + }}\pi {\rm{/2)}}$

= –asin($\omega $t + $\alpha $) ...(ii)

Given the two waves are acting in perpendicular direction with the same frequency and phase difference $\pi $/2.

From equations (i) and (ii),
x² + y² = a²
which represents the equation of a circle.

$f' = {{v - u} \over v}f;f'' = {{v + u} \over v}f$

Number of beats = $f'' - f' = {{2u} \over \lambda }$

Given, wave equation, $y=0.002 \sin (300 t-15 x)$

Mass density, $\mu=0.1 \mathrm{~kg} / \mathrm{m}$

Comparing the given wave equation with $y=a \sin (\omega t-k x)$, we have $\omega=300, k=15$

$\therefore$ Velocity of wave,

$v=\frac{\omega}{k}=\frac{300}{15}=20 \mathrm{~m} / \mathrm{s}$

Wave speed in string is also given by

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

where, $T=$ tension in the string and $\mu=$ mass per unit length of string

$\therefore 20=\sqrt{\frac{T}{0.1}} \Rightarrow 400=\frac{T}{0.1} \Rightarrow T=40 \mathrm{~N}$

Given, frequency of sound produced source $S_1$ and $S_2$ i.e, $f_1=f_2=500 \mathrm{~Hz}$

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

Beats $=3 \mathrm{~Hz}$

If $v_s$ be the speed of source, when first source is moving towards the observer and $f_1{ }^{\prime}$ be apparent frequency heard by the observer, then,

$f_1{ }^{\prime}=\frac{v}{v-v_s} \times f_1=\frac{330}{330-v_s} \times 500$

When second source is moving away from the apparent frequency heard by the observer, then

$f_2^{\prime}=\frac{v}{v+v_s} \times f_2=\frac{330}{330+v_s} \times 500$

$\begin{aligned} & \therefore \text { Beats }=f_1{ }^{\prime}-f_2{ }^{\prime}=3 \\ & \Rightarrow \frac{330}{330-v_s} \times 500-\frac{330}{330+v_s} \times 500=3 \\ & \Rightarrow \frac{500 \times 2 v_s \times 330}{330^2-v_s^2}=3 \end{aligned}$

Since, $v_s^2<<330^2$

$\Rightarrow \frac{500 \times 2 v_s \times 330}{330^2}=3 \Rightarrow v_s=0.99 \approx 1 \mathrm{~m} / \mathrm{s}$

Given, speed of sound, $v=330 \mathrm{~m} / \mathrm{s}$

Length of organ pipe, $l=1 \mathrm{~m}$

Frequency, $f=1000 \mathrm{~Hz}$

For open organ pipe,

Fundamental frequency, $f_0=\frac{v}{2 l}=\frac{330}{2 \times 1}=165 \mathrm{~Hz}$

$\therefore$ Number of tones present in the open organ pipe

$=\frac{f}{f_0}=\frac{1000}{165}=6.06 \simeq 6$

In fundamental mode, $l=2\left(\frac{\lambda}{4}\right)=\frac{\lambda}{2}$

$\Rightarrow \quad \lambda=2 l$ ..... (i)

Given, $l=100 \mathrm{~cm}, v=2.53 \mathrm{~kHz}=2.53 \times 10^3 \mathrm{~Hz}$

We know that,

$\begin{aligned} v & =v \lambda \\ & =v \times 2 l \quad \text{[from Eq. (i)]}\\ & =2.53 \times 10^3 \times 2 \times 100 \times 10^{-2} \\ & =5.06 \times 10^3 \mathrm{~m} / \mathrm{s} \\ & =5.06 \mathrm{~km} / \mathrm{s} \end{aligned}$

For closed organ pipe,

$f=\frac{(2 n+1) v}{4 l} \text { (where, } n=0,1,2, ...)$

According to question,

$\begin{aligned} \frac{(2 n+1) v}{4 l} & <1250 \\\\ 2 n+1 & <\frac{1250 \times 4 l}{v} \\\\ 2 n+1 & <\frac{1250 \times 4 \times 0.85}{340} \\\\ \Rightarrow \quad 2 n & <11.5 \\\\ n & < 5.75 \end{aligned}$

So $n=0,1,2,3, \ldots 5$.

So, we have 6 possibilities.

According to the property of persistence of hearing the impression of a sound heard perist on our mind for $\frac{1}{10} \mathrm{~s}$. To hear distinct beats, time interval between two successive beats should be greater than $\frac{1}{10} \mathrm{~s}$. Hence, difference in two frequencies must be less than 10.

To find the frequency of the reflected sound as heard by the driver, we can break the problem into two steps:

Calculate the Frequency Heard by the Hill:

When the car is moving towards the hill, the frequency of the sound heard by the hill ($f'$) changes due to the Doppler effect. The formula for this scenario (source moving towards a stationary observer) is:

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

where:

$ f = 600 \, \text{Hz} $ is the original frequency of the horn.

$ v = 330 \, \text{m/s} $ is the speed of sound in air.

$ v_0 = 30 \, \text{m/s} $ is the speed of the car (source).

Substituting the values, we have:

$ f' = 600 \left( \frac{330 + 30}{330} \right) = 600 \left( \frac{360}{330} \right) = 600 \times \frac{12}{11} = 654.54 \, \text{Hz}. $

Calculate the Frequency of the Reflected Sound Heard by the Driver:

The reflected sound now acts as a new source of frequency $ f' $ as calculated above. Since the driver is moving towards this new source (the hill), we apply the Doppler effect again:

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

Substituting the values:

$ f'' = 654.54 \left( \frac{330}{330 - 30} \right) = 654.54 \left( \frac{330}{300} \right) = 654.54 \times \frac{11}{10} = 720 \, \text{Hz}. $

Thus, the frequency of the reflected sound as heard by the driver is:

Option C: $720 \, \text{Hz}$.