Waves

We are given two waves:

$ x_1 = A \sin (\omega t + \frac{\pi}{6}) $ and $ x_2 = A \cos \omega t $

Step 1: Find the phase of each wave

For the first wave, $x_1$, the phase is the angle inside the sine function: $ \text{Phase of } x_1 (\phi_1) = \frac{\pi}{6} $

For the second wave, $x_2 = A \cos \omega t$, we can write cosine as sine: $ \cos \omega t = \sin \left(\omega t + \frac{\pi}{2}\right) $ So, $ x_2 = A \sin \left(\omega t + \frac{\pi}{2}\right) $ The phase of $x_2$ is: $ \text{Phase of } x_2 (\phi_2) = \frac{\pi}{2} $

Step 2: Find the phase difference

The phase difference between the waves is: $ \Delta \phi = \phi_2 - \phi_1 $

Substitute the values: $ \Delta \phi = \frac{\pi}{2} - \frac{\pi}{6} $

Find a common denominator and subtract: $ = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} $

So, the phase difference between the waves is $\frac{\pi}{3}$.

From above wave diagram,

Number of node, N = 6

and number of anti-node, A = 5

Speed of transverse wave, $v=\sqrt{\eta / \rho}$

where, $\quad \eta=$ modulus of elasticity $\rho=$ density

$\therefore(\mathrm{A}) \rightarrow(2)$

Speed of longitudinal wave in earth crust

$v=\sqrt{B+\frac{4}{3}\left(\frac{\eta}{\rho}\right)} $

$\therefore(\mathrm{B}) \rightarrow(1)$

Speed of longitudinal wave, $v=\sqrt{\lambda / \rho}$ where, $\lambda=$ wavelength of wave.

$\therefore(\mathrm{C}) \rightarrow(4)$

Ripples speed, $v=\sqrt{\frac{2 \pi T}{\lambda \rho}}$

where, $T=$ surface tension coefficient.

$\therefore(\mathrm{D}) \rightarrow(3)$

Given, frequency of wave, $f=350 \mathrm{~Hz}$

Amplitudes of wave $a_1, a_2$ is $0.3 \mathrm{~mm}$ and $0.4 \mathrm{~mm}$, respectively.

Path difference $(A P-B P=\Delta x)=25 \mathrm{~cm}$

Speed of sound, $v=350 \mathrm{~ms}^{-1}$

As, we know that,

$\begin{aligned} & \Delta \phi=\frac{2 \pi}{\lambda} \Delta x \\ & \text { and (wavelength) } \lambda=\frac{\text { (speed) } v}{\text { (frequency) } f} \\ & =\frac{350}{350}=1 \mathrm{~m}=100 \mathrm{~cm} \\ & \Delta \phi=\frac{2 \pi}{100} \times 25=\pi / 2 \\ \end{aligned}$

As, resultant amplitude,

$\begin{aligned} a_R & =\sqrt{a_1^2+a_2^2+2 a_1 a_2 \cos \phi} \\ \Rightarrow \quad a_R & =\sqrt{0.3^2+0.4^2+2 \times 0.3 \times 0.4 \cos \pi / 2} \\ & =\sqrt{9 / 100+16 / 100} \\ & =\sqrt{25 / 100} \\ & =5 / 10=0.5 \mathrm{~mm} \end{aligned}$

Given, frequency heard by first truck driver,

$ v=495 \mathrm{~Hz} $

Velocity of sound, $v_s=u$

Velocity of t rucks, $v_1=v_2=0.1 u$

In the first case, when two trucks are approaching each other:

The frequency heard by second truck driver,

$ \begin{aligned} v_1 & =\left(\frac{v_s+v_2}{v_s-v_1}\right) v=\left(\frac{u+0.1 u}{u-0.1 u}\right) 495 \\ & =\frac{1.1}{0.9} \times 495=605 \mathrm{~Hz} \end{aligned} $

In the second case, when two trucks are receding each other:

The frequency heard by second truck driver,

$ \begin{aligned} v_2 & =\left(\frac{v_s-v_2}{v_s+v_1}\right) v=\left(\frac{u-0.1 u}{u+0.1 u}\right) 495 \\ & =\frac{0.9}{1.1} \times 495=405 \mathrm{~Hz} \\ v_1 & -v_2=605-405=200 \mathrm{~Hz} \end{aligned} $

Phase of given wave is

$ \phi=a x^2+b t^2+2 \sqrt{a b} x t $

Differentiating with respect to time,

$ \begin{aligned} & \frac{d \phi}{d t}=0=2 a x \frac{d x}{d t}+2 b t+2 \sqrt{a b}\left(t \cdot \frac{d x}{d t}+x\right) \\ & \Rightarrow(2 a x+2 \sqrt{a b} t) \frac{d x}{d t}=-(2 \sqrt{a b} x+2 b t) \end{aligned} $

So, wave velocity is

$ v=\frac{d x}{d t}=\frac{-(2 \sqrt{a b} \cdot x+2 b t)}{(2 a x+2 \sqrt{a b} \cdot t)}=-\sqrt{\frac{b}{a}} \mathrm{~ms}^{-1} $

According to Doppler's effect, Observed frequency of sound reflected from building is given by

$ f^{\prime}=f\left(\frac{v+v_b}{v-v_b}\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Here, $v=$ speed of sound $=340 \mathrm{~m} / \mathrm{s}$, $v_b=$ speed of bus $=7.2 \mathrm{~km} / \mathrm{h}=20 \mathrm{~ms}^{-1}$ and $f=$ original frequency produced by horn

             $ \text { = } 1.7 \mathrm{kHz} $

Hence, putting these value in Eq. (i), we get

Observed frequency by bus driver,

$ \begin{aligned} f^{\prime} & =1.7\left(\frac{340+20}{340-20}\right)=1.9125 \mathrm{kHz} \\ & \approx 1.8 \mathrm{kHz} \end{aligned} $

In first resonance, length of air column $ = {\lambda \over 4}$.

So, ${l_1} + e = {\lambda \over 4}$ or $11 \times 4 + 4e = \lambda $

So, speed of sound is

$ \Rightarrow v = {f_1}\lambda = 512(44 + 4e)$ ...... (i)

And in second case,

$l{'_1} + e = {{\lambda '} \over 4}$ or $27 \times 4 + 4e = \lambda '$

$ \Rightarrow v = {f_2}\lambda ' = 256(108 + 4e)$ ..... (ii)

Dividing both Eqs. (i) and (ii), we get

$1 = {{512(44 + 4e)} \over {256(108 + 4e)}} \Rightarrow e = 5$ cm

Substituting value of e in Eq. (i), we get

Speed of sound $v = 512(44 + 4e)$

$ = 512(44 + 4 \times 5)$

$ = 512 \times 64$ cm s^$-$1 = 327.68 ms^$-$1 $\approx$ 328 ms^$-$1