Waves

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