Waves

Given:

Both source and observer are moving towards each other.

Each moves at 10% of the speed of sound.

$ v_s = 0.1v, \quad v_o = 0.1v $

where $ v $ = speed of sound.

Formula (for Doppler effect):

When both are moving towards each other, the apparent frequency $ f' $ heard by the observer is

$ f' = f \frac{v + v_o}{v - v_s} $

Substituting given values:

$ f' = f \frac{v + 0.1v}{v - 0.1v} = f \frac{1.1v}{0.9v} = f \times \frac{1.1}{0.9} $

$ \frac{1.1}{0.9} = 1.222... $

So,

$ f' = 1.222 f $

Percentage change in frequency:

$ \%\ \text{increase} = (1.222 - 1) \times 100 = 22.2\% $

✅ Final Answer:

$ \boxed{22.2\%} \text{ (Option C)} $

Velocity of sound wave

$ v=\frac{\text { Distance }}{\text { Time }}=\frac{600}{2}=300 \mathrm{~m} / \mathrm{s} $

∴ Wavelength, $\lambda=\frac{v}{f}=\frac{300}{500}=0.6 \mathrm{~m}$

Number of waves between the points $X$ and $Y=\frac{\text { Distance }}{\text { Wavelength }}=\frac{600}{0.6}=1000$

$ \begin{aligned} &\text { From the given wave equation }\\ &\begin{aligned} k & =5 \times 10^{-3} \mathrm{~cm}^{-1}=5 \times 10^{-1} \mathrm{~m}^{-1} \\ & =0.5 \mathrm{~m}^{-1} \\ w & =20 \mathrm{rad} / \mathrm{s} \\ \therefore \quad v & =\frac{\omega}{k}=\frac{20}{5 \times 10^{-1}}=40 \mathrm{~m} / \mathrm{s} \\ \therefore \quad \mu & =\frac{m}{l}=\frac{3 \times 10^{-3}}{0.8} \\ & =3.75 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \\ \therefore \text { Since, } v & =\sqrt{\frac{T}{\mu}} \\ \Rightarrow \quad v^2 & =\frac{T}{\mu} \\ \Rightarrow \quad T & =\mu v^2=3.75 \times 10^{-3} \times 40^2 \\ & =3.75 \times 10^{-3} \mathrm{~N}=6 \mathrm{~N} \end{aligned} \end{aligned} $

Standard equation of a travelling wave is,

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

Given, $y=0.5 \sin (70.1 x-10 \pi t)$

$\therefore \omega=$ angular frequency $=10 \pi$

But $\omega=2 \pi f, f=$ frequency

$ \Rightarrow f=\frac{\omega}{2 \pi}=\frac{10 \pi}{2 \pi}=5 \mathrm{~Hz} $

Phase difference between waves is

$ \begin{aligned} \phi_1-\phi_2 & =\left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right)-\left(\omega t-\frac{2 \pi x}{\lambda}\right) \\ & =\phi \end{aligned} $

As, relation of phase and path difference is,

$ \frac{\Delta \phi}{2 \pi}=\frac{\Delta L}{\lambda} $

$\Rightarrow$ Path difference, $\Delta L=\frac{\lambda \phi}{2 \pi}$.

$ \begin{aligned} &\text { } y_1=3 \sin 250 \pi t\\ &\begin{array}{ll} \therefore & \omega_1=250 \pi \mathrm{rad} / \mathrm{s} \\ & y_2=2 \mathrm{sin} 260 \pi t \\ \therefore & \omega_2=260 \pi \mathrm{rad} / \mathrm{s} \\ \therefore & f_1=\frac{\omega_1}{2 \pi}=\frac{250 \pi}{2 \pi}=125 \mathrm{~Hz} \\ & f_2=\frac{\omega_2}{2 \pi}=\frac{260 \pi}{2 \pi}=130 \mathrm{~Hz} \\ & T=\frac{1}{f_2-f_1} \\ & =\frac{1}{130-125}=\frac{1}{5}=0.2 \mathrm{~s} \end{array} \end{aligned} $

In a closed organ pipe, the number of nodes formed in the $n$th harmonic is equal to $\left(\frac{n+1}{2}\right)$ if $n$ is odd.

Thus, for the fifth $(n=5)$ harmonic, the number of nodes $=\frac{5+1}{2}=3$

and for the ninth harmonic ( $n=9$ ), the number of nodes $=\frac{9+1}{2}=5$

Given:

A stretched wire has a fundamental frequency $ f $.

When it is divided into 3 segments, their fundamental frequencies are $ f_1, f_2, $ and $ f_3 $.

Tension $ T $ is constant.

We need to find the relation among $ f, f_1, f_2, f_3 $.

Step 1: Formula for fundamental frequency of a stretched string

For a string under tension $ T $, length $ L $, and linear density $ \mu $:

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

Step 2: Frequencies of the three parts

Suppose the original length is divided into three parts of lengths $ L_1, L_2, L_3 $, so that

$ L = L_1 + L_2 + L_3 $

Then their fundamental frequencies are

$ f_1 = \frac{1}{2L_1}\sqrt{\frac{T}{\mu}}, \quad f_2 = \frac{1}{2L_2}\sqrt{\frac{T}{\mu}}, \quad f_3 = \frac{1}{2L_3}\sqrt{\frac{T}{\mu}} $

Step 3: Express $ L_i $ in terms of $ f_i $

From the above,

$ L_i = \frac{1}{2f_i}\sqrt{\frac{T}{\mu}} $

and

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

Step 4: Use $ L = L_1 + L_2 + L_3 $

$ \frac{1}{2f}\sqrt{\frac{T}{\mu}} = \frac{1}{2}\sqrt{\frac{T}{\mu}} \left(\frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}\right) $

Cancel the common factor $ \frac{1}{2}\sqrt{\frac{T}{\mu}} $:

$ \boxed{\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}} $

✅ Correct Option: (C)

$ \boxed{\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}} $

Mass per unit length,

$ \mu=\frac{M}{l}=\frac{5 \times 10^{-3}}{0.81} $

$\therefore$ Speed of transverse wave

$ \begin{aligned} v & =\sqrt{\frac{T}{\mu}}=\sqrt{\frac{50}{\frac{5 \times 10^{-3}}{0.81}}} \\ & =\sqrt{\frac{50 \times 0.81}{5 \times 10^{-3}}}=90 \mathrm{~m} / \mathrm{s} \end{aligned} $

$y=0.7 \sin \left(\frac{7 \pi}{4} x\right) \cos (350 \pi t)$

Comparing with equation of stationary wave,

$y=2 A \sin (k x) \cos \omega t$

We get, $k=\frac{7 \pi}{4}, \omega=350 \pi$

$\therefore$ Speed of stationary wave,

$ v=\frac{\omega}{k}=\frac{350 \pi}{\frac{7 \pi}{4}}=200 \mathrm{~m} / \mathrm{s} $

$ \begin{aligned} \text { } f_1 & =\frac{v}{\lambda_1}=\frac{330}{0.99}=333.33 \mathrm{~Hz} \\ f_2 & =\frac{v}{\lambda_2}=\frac{330}{1}=330 \mathrm{~Hz} \\ \therefore f_{\text {beat }} & =f_1-f_2=333.33-330=3.33 \mathrm{~Hz} \\ \therefore \quad t & =\frac{\text { Total number of beat }}{f_{\text {beat }}}=\frac{10}{3.33}=3 \mathrm{~s} \end{aligned} $

Here’s how to find the closed‐pipe length $L_c$:

For a closed pipe (one end closed), only odd harmonics appear:

$ f_m = \frac{m\,v}{4\,L_c}\quad (m=1,3,5,\dots) $

For an open pipe (both ends open), all harmonics appear:

$ f_n = \frac{n\,v}{2\,L_o}\quad (n=1,2,3,\dots) $

Equate the 5th harmonic of the closed pipe to the 3rd harmonic of the open pipe:

$ \frac{5\,v}{4\,L_c} \;=\; \frac{3\,v}{2\,L_o} $

Solve for $L_c$:

$ \frac{5}{4\,L_c} = \frac{3}{2\,L_o} \;\;\Longrightarrow\;\; L_c = \frac{5}{6}\,L_o = \frac{5}{6}\times 72\;\text{cm} = 60\;\text{cm} $

Answer: 60 cm (Option A).

To determine the new fundamental frequency when changing the configuration of the pipe, follow these steps:

Initial Conditions

Length of the pipe (L): The total length of the open pipe.

Speed of sound (v): A constant parameter in this context.

Open Pipe Fundamental Frequency

For an open pipe, the fundamental frequency ($f_0$) is given by:

$ f_0 = \frac{v}{2L} $

Given that $f_0 = 100 \, \text{Hz}$, we have:

$ 100 = \frac{v}{2L} \quad \Rightarrow \quad v = 200L $

Closed Pipe Adjustment

When the pipe's bottom end is closed and $\frac{1}{3}$ of the pipe is filled with water, the air column effectively resonating is:

$ L' = L - \frac{L}{3} = \frac{2L}{3} $

Closed Pipe Fundamental Frequency

For a closed pipe, the fundamental frequency ($f_C$) is calculated using:

$ f_C = \frac{v}{4L'} $

Substitute $L'$ into the formula:

$ f_C = \frac{v}{4 \times \frac{2L}{3}} = \frac{3v}{8L} $

Plug in the value of $v$ from the earlier expression:

$ f_C = \frac{3 \times 200L}{8 \times L} = \frac{600}{8} = 75 \, \text{Hz} $

Thus, the new fundamental frequency when the pipe is closed and partly filled with water is 75 Hz.

The pipe is closed $f=\frac{\nu}{\lambda}$

$ \therefore \quad f_a=\frac{v}{4 L} $

The pipe is open

So, $f=\frac{v}{\lambda}$

Here, $L=\lambda$

$ f_b=\frac{v}{L} $

The pipe is open, $f=\frac{v}{\lambda}$

The pipe is closed

So, $f=\frac{v}{\lambda}$

Here, $L=\frac{3 \lambda}{4} \Rightarrow f_d=\frac{3 v}{4 L}$

Now, $f_a: f_b: f_c: f_d$

$ \frac{v}{4 L}: \frac{v}{L}: \frac{v}{2 L}: \frac{3 v}{4 L}=1: 4: 2: 3 $

When a wave transitions from a denser medium to a rarer medium, the property that remains constant is the frequency.

Frequency is determined by the source of the wave and remains unchanged when the wave passes from one medium to another.

Let's define:

$ f = 1000 \, \text{Hz} $: the original frequency of the horn.

$ v = 340 \, \text{m/s} $: the speed of sound.

$ v_c $: the speed of the car.

Apparent Frequency Formulas

Approaching Observer:

$ f_1 = f \left( \frac{v}{v - v_c} \right) $

Receding from Observer:

$ f_2 = f \left( \frac{v}{v + v_c} \right) $

Frequency Ratio Equation

Given $ \frac{f_1}{f_2} = \frac{11}{9} $, we equate and solve:

$ \frac{f \left( \frac{v}{v - v_c} \right) }{f \left( \frac{v}{v + v_c} \right) } = \frac{v + v_c}{v - v_c} = \frac{11}{9} $

Solving for $ v_c $

Starting from:

$ \frac{11}{9} = \frac{340 + v_c}{340 - v_c} $

Cross-multiplying gives:

$ 11(340 - v_c) = 9(340 + v_c) $

Simplifying:

$ 3740 - 11v_c = 3060 + 9v_c $

$ 680 = 20v_c $

$ v_c = \frac{680}{20} = 34 \, \text{m/s} $

Thus, the speed of the car is $ 34 \, \text{m/s} $.

Given:

Frequency, $ v_n = 1.65 \, \text{kHz} = 1650 \, \text{Hz} $

Velocity of sound, $ v = 330 \, \text{m/s} $

Length of pipe, $ L = 30 \, \text{cm} = 0.3 \, \text{m} $

The frequency of the $ n $th harmonic in an open pipe is calculated using the formula:

$ v_n = \frac{n v}{2 L} $

Substitute the given values:

$ 1650 = \frac{n \times 330}{2 \times 0.3} $

Solving for $ n $:

$ 1650 = \frac{n \times 330}{0.6} $

$ 1650 \times 0.6 = n \times 330 $

$ 990 = n \times 330 $

$ n = \frac{990}{330} = 3 $

Thus, the pipe resonates at the 3rd harmonic.

When the frequency of a wave is increased by 25%, the wavelength changes as follows (assuming the medium remains unchanged).

The speed of the wave remains the same because the medium does not change:

$ v = \text{constant} $

Therefore, according to the wave equation:

$ f_1 \lambda_1 = f_2 \lambda_2 $

If the initial frequency $ f_1 $ is increased by 25%, the new frequency $ f_2 $ becomes:

$ f_2 = 1.25 \times f_1 $

Substituting this into the wave equation:

$ f_1 \lambda_1 = (1.25 \times f_1) \lambda_2 $

Solving for the new wavelength $ \lambda_2 $:

$ \lambda_2 = \frac{\lambda_1}{1.25} = 0.8 \lambda_1 $

The change in wavelength $\Delta \lambda$ is calculated as:

$ \Delta \lambda = \frac{\lambda_2 - \lambda_1}{\lambda_1} \times 100 = \frac{0.8 \lambda_1 - \lambda_1}{\lambda_1} \times 100 $

Simplifying this gives:

$ \Delta \lambda = \frac{-0.2 \lambda_1}{\lambda_1} \times 100 = -20\% $

Thus, the wavelength decreases by 20%.

To determine the percentage increase in tension required to raise the wave speed by 20%, we start by examining the given relationship between wave speed and tension.

Step-by-Step Solution

Calculate the New Wave Speed:

Given that the original wave speed is $ v_1 $ and we need to increase it by 20%, the new speed $ v_2 $ is:

$ v_2 = v_1 + 20\% \text{ of } v_1 = v_1 + \frac{v_1}{5} = \frac{6v_1}{5} $

Relate Wave Speed to Tension:

The speed of a wave on a string is given by:

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

This implies:

$ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} $

Squaring both sides gives:

$ \frac{T_2}{T_1} = \left(\frac{v_2}{v_1}\right)^2 $

Substitute $ v_2 $ and $ v_1 $:

$ \frac{T_2}{T_1} = \left(\frac{\frac{6v_1}{5}}{v_1}\right)^2 = \left(\frac{6}{5}\right)^2 = \frac{36}{25} $

Find the Required Increase in Tension:

$ \frac{T_2 - T_1}{T_1} = \frac{T_2}{T_1} - 1 = \frac{36}{25} - 1 = \frac{11}{25} $

Convert to Percentage:

$ \frac{T_2 - T_1}{T_1} \times 100\% = \frac{11}{25} \times 100\% = 44\% $

Thus, the tension needs to be increased by 44% to achieve a 20% increase in wave speed.

To determine the frequency of string $ A $, we start with the following information:

Initial beat frequency is $ 4 $ beats per second.

Frequency of string $ B $, $ f_B = 480 \, \text{Hz} $.

The relation for beat frequency: $|f_A - f_B| = 4$.

Given that the tension on string $ A $ is increased, the frequency of $ A $ will increase, leading to an increase in the beat frequency. The new beat frequency becomes $ 7 $ beats per second.

The vibration frequency of a string is determined by the formula:

$ f = \frac{n}{2L} \cdot \sqrt{\frac{T}{\mu}} $

Where:

$ n $ is the mode of vibration.

$ L $ is the length of the string.

$ T $ is the tension in the string.

$ \mu $ is the linear mass density.

Since the tension in string $ A $ is increased, we know the frequency of $ A $ increases. Consequently, the following must be true for the beat frequency to increase:

$ f_A - f_B = 4 \quad \text{or} \quad f_B - f_A = 4 $

However, since the frequency of $ A $ increases and the beat frequency becomes 7, it implies that:

$ f_A - f_B = 7 $

Given:

Initially, $ f_A - 480 = 4 $, thus $ f_A = 484 \, \text{Hz} $.

Therefore, as $ f_A $ increases past 480 Hz and still satisfies the condition where the beat frequency becomes $ 7 $, this confirms that initially:

$ f_A = 484 \, \text{Hz} $

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

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

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,

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