Waves

To find the frequency of the plane progressive wave given by the equation $y = 2 \cos 2 \pi(330 \mathrm{t} - x) \mathrm{m}$, we start by analyzing the general form of a wave equation.

The general form of a wave equation is:

where:

• $A$ is the amplitude of the wave.
• $f$ is the frequency of the wave.
• $t$ is the time variable.
• $k$ is the wave number, defined as $\frac{2 \pi}{\lambda}$ where $\lambda$ is the wavelength.
• $x$ is the spatial variable.
• $\phi$ is the phase constant.

By comparing the given wave equation with the general form, we have:

We observe that the term $2 \pi(330t - x)$ corresponds to $2 \pi ft - kx$ in the general form.

From this, it is clear that:

• $2 \pi ft \rightarrow 2 \pi \cdot 330 t$
• Therefore, $f = 330$ Hz

So, the frequency of the wave is 330 Hz.

Thus, the correct answer is:

Option C: 330 Hz

$\begin{aligned} & \mathrm{f}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1} \\ & \mathrm{f}_1=\mathrm{f}_2 \\ & \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{\mathrm{v}}{\mathrm{L}_2} \\ & \Rightarrow \mathrm{L}_2=4 \mathrm{~L}_1 \\ & 60=4 \times \mathrm{L}_1 \\ & \mathrm{~L}_1=15 \mathrm{~cm} \end{aligned}$

$C_1=\sqrt{\frac{T}{\mu}}, C_2=\sqrt{\frac{T}{4 \mu}}=\frac{C_1}{2}$

For node at O :

$\begin{aligned} & \mathrm{L}=\frac{\mathrm{n} \lambda_1}{2}, 2 \mathrm{~L}=\frac{\mathrm{m} \lambda_2}{2}(\mathrm{n}, \mathrm{m} \text { are integers }) \\ & \lambda_1=\frac{2 \mathrm{~L}}{\mathrm{n}}, \lambda_2=\frac{4 \mathrm{~L}}{\mathrm{~m}} \\ & \frac{\mathrm{C}_1}{\lambda_1}=\frac{\mathrm{C}_2}{\lambda_2} \\ & \Rightarrow \frac{\mathrm{C}_1}{\frac{2 \mathrm{~L}}{\mathrm{n}}}=\frac{\frac{\mathrm{C}_1}{2}}{\frac{2 \mathrm{~L}}{\mathrm{~m}}} \\ & \Rightarrow 4 \mathrm{n}=\mathrm{m} \end{aligned}$

For minimum frequency, $\mathrm{n}=1, \mathrm{~m}=4$

$\therefore v_{\min }=\frac{\mathrm{C}_1 \times 1}{2 \mathrm{~L}}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}=v_0$

The string will look like

Total no. of nodes $=6$ including the end nodes

For antinode at O :

$\begin{aligned} & \mathrm{L}=(2 \mathrm{n}+1) \frac{\lambda_1}{4} ; 2 \mathrm{~L}=(2 \mathrm{n}+1) \frac{\lambda_2}{4} \quad \text { ( } \mathrm{n}, \mathrm{m} \text { are integers) } \\ & \lambda_1=\frac{4 \mathrm{~L}}{(2 \mathrm{n}+1)} ; \lambda_2=\frac{8 \mathrm{~L}}{(2 \mathrm{~m}+1)} \\ & \frac{C_1}{\lambda_1}=\frac{C_2}{\lambda_2} \\ & \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{\lambda_1}{\lambda_2} \\ & 2=\frac{\frac{4 L}{(2 n+1)}}{\frac{8 \mathrm{~L}}{(2 m+1)}} \\ \end{aligned}$

$4=\frac{(2 m+1)}{(2 n+1)} \Rightarrow \text { even }=\frac{\text { odd }}{\text { odd }} \Rightarrow \text { This node is not possible }$

Given,

Length of the wire, $ L=1 \mathrm{~m} $

Elastic strain, $ \varepsilon=1 \%=0.01 $

Density of metal, $ \rho=8000 \mathrm{~kg} / \mathrm{m}^{3} $

Young's modulus, $ Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m} $

We know that, $ Y=\frac{\sigma \text { (stress) }}{\varepsilon \text { (strain) }} $

$ \begin{array}{l} \sigma=Y \cdot \varepsilon=2 \times 10^{11} \times 0.01 \\\\ \sigma=2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} \end{array} $

We also new that,

$ \sigma=\frac{T}{A} \Rightarrow T=\sigma \times A $

Cross-sectional area is not given, so we consider mass per unit length $ \mu $ of the wire,

$ \mu=\rho \cdot A $

Putting in Eq. (i), we get

$ T=\frac{\sigma \cdot \mu}{\rho} $

The fundamental frequency of the transvèrse waves in the metal wire is,

$ \begin{array}{l} f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \\\\ f=\frac{1}{2 \times 1} \sqrt{\frac{\sigma \cdot \mu}{\rho \cdot \mu}}=\frac{1}{2} \sqrt{\frac{\sigma}{\rho}} \\\\ f=\frac{1}{2} \sqrt{\frac{2 \times 10^{9}}{8000}}=\frac{1}{2} \times \sqrt{\frac{10^{6}}{4}} \\\\ f=\frac{1}{4} \times 10^{3}=\frac{1000}{4} \\\\ f=250 \mathrm{~Hz} \end{array} $

Given,

Beat frequency, $ f_{\text {beat }}=6 \mathrm{~Hz} $

Length of the shorter pipe, $ L_{1}=150 \mathrm{~cm} $

$ =1.5 \mathrm{~m} $

Speed of sound in air, $ v=336 \mathrm{~m} / \mathrm{s} $

For a closed loop, fundamental frequency is given by,

$ f=\frac{v}{4 L} $

For the shorter pipe $ L_{1} $,

$ f_{1}=\frac{336}{4 \times 1.5}=\frac{336}{6}=56 \mathrm{~Hz} $

Let the length of longer pipe be $ L_{2} $ and frequency be $ f_{2} $.

Since the pipes produces 6 beats per second when sound together:

$ \left|f_{1}-f_{2}\right|=6 $

There are two possible cases for $ f_{2} $,

Case (i) $ f_{2}=f_{1}+6=56+6=62 \mathrm{~Hz} $

Case (ii) $ f_{2}=f_{1}-6=56-6=50 \mathrm{~Hz} $

So, from Eq. (i),

Case (i) $ f_{2}=\frac{v}{4 L_{2}} \Rightarrow 62=\frac{336}{4 L_{2}} $

$ \Rightarrow \quad L_{2}=\frac{336}{4 \times 62} $

$ L_{2}=1.35 \mathrm{~m}=135 \mathrm{~cm} $ (Not used because it is less than 4)

Case (ii) $ f_{2}=\frac{v}{4 L_{2}} \Rightarrow 50=L_{2}=\frac{336}{4 L_{2}} $

$ \Rightarrow \quad L_{2}=\frac{336}{4 \times 50} $

$ f_{2}=1.68 \mathrm{~m}=168 \mathrm{~cm} $

So, the length of the longer pipe that results in the production of 6 beats per second when sounded with the shorter pipe is 168 cm .

Given:

Path difference ($\Delta x$): 50 cm = 0.5 m

Phase difference ($\phi$): $1.8 \pi$

Velocity of sound ($v$): 340 m/s

We are to find the frequency of the sound wave.

First, use the relationship between phase difference, path difference, and wavelength:

$ \phi = \frac{2 \pi}{\lambda} \times \Delta x $

Substitute the given values:

$ 1.8 \pi = \frac{2 \pi}{\lambda} \times 0.5 $

Solve for $\lambda$ (wavelength):

$ 1.8 \pi = \frac{\pi}{\lambda} $

$ \lambda = \frac{1}{1.8} \, \text{m} $

Next, use the equation relating velocity, frequency, and wavelength:

$ v = f \times \lambda $

Substitute the known values:

$ 340 = f \times \frac{1}{1.8} $

Solve for $f$ (frequency):

$ f = 340 \times 1.8 $

$ f = 612 \, \text{Hz} $

Thus, the frequency of the sound wave is 612 Hz.

Given:

Frequency of source, $ f_{0} = 102 \, \text{Hz} $

Source velocity, $ v_{S} = 0 \, \text{m/s} $

Observers' Motion:

Each observer is moving with a velocity that is 10% of the speed of sound, i.e., $ v_{0} = 0.1v $.

Doppler Effect Calculation:

For each observer moving away from the source, the frequency heard is given by:

$ f_{\text{observer}} = f_{0} \left( \frac{v - v_{0}}{v} \right) $

where $ v $ is the speed of sound in the medium.

Frequency Calculation for Both Observers:

Since both observers are moving away from the source at the same speed and in opposite directions, the formula for frequency remains consistent for both:

$ f_{\text{observer 1}} = f_{0} \left( \frac{v - 0.1v}{v} \right) = f_{0} \left( \frac{0.9v}{v} \right) = 0.9 f_{0} $

$ f_{\text{observer 2}} = f_{0} \left( \frac{v - 0.1v}{v} \right) = f_{0} \left( \frac{0.9v}{v} \right) = 0.9 f_{0} $

Ratio of Frequencies:

The ratio of the frequencies heard by the two observers is:

$ \frac{f_{\text{observer 1}}}{f_{\text{observer 2}}} = \frac{0.9 f_{0}}{0.9 f_{0}} = 1:1 $

Thus, both observers hear the same frequency, resulting in a ratio of 1:1.

Fundamental frequency of open pipe

$ f_{O}=\frac{v}{2 L} $

Fundamental frequency of closed pipe

$ \begin{aligned} f_{C} & =\frac{v}{4 L} \\\\ \frac{v}{2 L}-\frac{v}{4 L} & =100 \mathrm{~Hz} \\\\ \frac{v}{4 L} & =100 \mathrm{~Hz} \end{aligned} $

(given)

Now, 2nd harmonic of open pipe

$ \begin{aligned} f & =\frac{2 v}{2 L} \end{aligned}\left(f_{n}=\frac{n v}{2 L}\right) $

3rd harmonic of closed pipe

$ (2 n-1) \frac{v}{4 L}=\frac{5 v}{4 L} $

Difference, $ \frac{5 v}{4 L}-\frac{v}{L}=\frac{v}{4 L} $

$ \frac{v}{4 L}=100 \mathrm{~Hz} $

$ y_{1}=5 \sin 400 \pi t, y_{2}=8 \sin 408 \pi t $

Compare with standard wave equation

$ \begin{aligned} y & =A \sin \omega t \\\\ \omega_{1} & =400 \pi, \omega_{2}=408 \pi \end{aligned} $

We know that, $ \omega=2 \pi f $

$ \begin{aligned} \omega_{1} & =2 \pi f_{1}=400 \pi \\\\ f_{1} & =200 \mathrm{~Hz} \\\\ \omega_{2} & =2 \pi f_{2}=408 \pi \\\\ f_{2} & =204 \mathrm{~Hz} \end{aligned} $

Number of beats/second

$ f_{2}-f_{1}=204-200=4 \mathrm{~Hz} $

Now, number of beats per minute

$ \begin{array}{l} =4 \times 60 \\\\ =240 \text { beats/minute } \end{array} $

Give velocity of train, $v_t=108 \mathrm{kmph}$

$ =108 \times \frac{5}{18}=30 \mathrm{~m} / \mathrm{s} $

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

From Doppler's effect,

$ f_0=\frac{v_s+v_{\text {obberver }}}{v_s+v_{\text {train }}} \times f_s $

When train is going way

$ \begin{aligned} \Rightarrow \quad f_1 & =\frac{330}{330+30} f_s \\\\ f_1 & =\frac{330}{360} f_s \\\\ f_1 & =0.9166 f_s \end{aligned} $

When train is closing in

$ \begin{aligned} f_2 & =\frac{330}{330-30} f_s \\\\ & =11 f_s \end{aligned} $

$\therefore \%$ change in frequency observed

$ \begin{aligned} & =\frac{11 f_s-0.9166 f_s}{11 f_s} \times 100 \% \\\\ & =0.1667 \times 100 \% \\\\ & =16.67 \% \end{aligned} $

Number of beats produced between $n$ and $n-1$

$ \begin{aligned} & =n-(n-1) \\\\ & =1 \end{aligned} $

Number of beats produced between $n$ and $n+1$

$ \begin{aligned} & =(n+1)-(n) \\\\ & =1 \end{aligned} $

Number of beats produced between $n+1$ and $n-1$

$ \begin{aligned} & =(n+1)-(n-1) \\\\ & =2 \end{aligned} $

Total number of beats produced

$ =1+1+2=4 $

Number of distinguishable beat per second,

$ (n+1)-(n-1)=2 $

$ \begin{aligned} \mathrm{v}_1 & =\mathrm{v}_0\left[\frac{v-v_0}{v-v_t}\right] \\\\ & =400\left[\frac{335-0}{335-15}\right]=\frac{400 \times 335}{320} \mathrm{~Hz} \end{aligned} $

$\begin{aligned} v_2 & =v_0\left[\frac{v-v_0}{v-v_1}\right] \\\\ & =400\left[\frac{335-0}{335-15}\right]=\frac{400 \times 335}{320} \mathrm{~Hz} \\\\ v_1 & -v_2=0\end{aligned}$

Given,

Speed of a transverse wave in a

stretched spring $A=v$

In a stretched spring $B=$ ?

The speed of transverse wave in a

stretched spring, $v_A=\sqrt{\frac{T}{\mu}}$

where, $v=$ speed of the wave

$T=$ tension in the spring

$\mu=$ linear mass density of the spring

Speed of wave in string $B$,

$ \begin{aligned} & v_B=\sqrt{\frac{T}{\mu_B}} \\\\ & v_B=\sqrt{\frac{T}{1.02 \mu_A}} \end{aligned} $

Since, tension $T$ is same for both strings, it cancels out

$ \begin{aligned} & v_B=\sqrt{\frac{1}{1.02} \cdot \frac{T}{\mu_A}} \\\\ & v_B=\frac{1}{\sqrt{1.02}} \cdot v \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} $

Using the Doppler effect formula:

$f = f_0\left(\frac{c + v_0}{c + v_s}\right)$

where

• $f$ is the observed frequency (frequency heard by the passenger in car Q)
• $f_0$ is the source frequency (400 Hz)
• $c$ is the speed of sound (360 m/s)
• $v_0$ is the velocity of the observer (passenger in car Q) relative to the medium (air) in the direction of the source (40 m/s)
• $v_s$ is the velocity of the source (car P) relative to the medium (air) in the direction of the observer (20 m/s, since it's moving in the same direction as car Q)

Plugging in the values:

$f = 400\left(\frac{360 + 40}{360 + 20}\right)$

$f = 400\left(\frac{400}{380}\right)$

$f = 421$

So, the observed frequency is 421 Hz.

We can write the given equation as:

$y = \sqrt{(\sin\omega t)^2 + (\cos\omega t)^2} \cos\left(\omega t - \arctan\frac{\sin\omega t}{\cos\omega t}\right)$

Using the identity $\sin^2\theta + \cos^2\theta = 1$, we get:

$y = \sqrt{1 + \sin 2\omega t} \cos\left(\omega t - \frac{\pi}{4}\right)$

The amplitude of the motion is the maximum value of $|y|$, which occurs when $\sin 2\omega t = 1$, i.e., at $t = \frac{\pi}{4\omega} + \frac{n\pi}{\omega}$, where $n$ is an integer. Substituting this value of $t$ in the above equation, we get:

$|y_{\text{max}}| = \sqrt{1 + \sin \frac{\pi}{2}} = \sqrt{2}$

Therefore, the amplitude of the motion is $\sqrt{2}$

The phenomenon of frequency change due to relative motion between a source and an observer is called Doppler effect. However, if the observer is at rest relative to the source (as in this case where the passenger is inside the train), then the frequency heard by the observer is the same as the frequency produced by the source.

This is because the relative velocity between the source (whistle) and the observer (passenger in the train) is zero. The Doppler effect only applies when there is a relative velocity between the source and the observer.

Therefore, the frequency heard by the passenger inside the train is the same as the frequency of the whistle, i.e., 400 Hz.

So, the correct answer is:

400 Hz.

Given,

Length of string, $r=50 \mathrm{~cm}=0.5 \mathrm{~m}$

Angular speed $=40 \mathrm{rad} / \mathrm{s}$

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

Using formula,

$ v=r \omega=0.5 \times 40=20 \mathrm{~m} / \mathrm{s} $

Now maximum frequency is given

$ \begin{aligned} f_{\max } & =f_{\text {standard }} \times\left[\frac{v}{v_s-v}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ v & =\text { speed of source's sound } \\ v_s & =\text { speed of sound } \\ f_{\min } & =f_{\text {standard }} \times\left[\frac{v}{v_s+v}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i) and (ii)

$ \frac{f_{\max }}{f_{\min }}=\frac{v_s+v}{v_s-v}=\frac{340+20}{340-20}=\frac{360}{320}=\frac{9}{8} $

Hence, the ratio of maximum to minimum frequency is $9: 8$

Given, Acceleration due to gravity $=g=10 \mathrm{~m} / \mathrm{s}^2$ When lift goes up with acceleration $\left(a_1\right)= 2.1 \mathrm{~m} / \mathrm{s}^2$

Lift goes down with acceleration $\left(\mu_2\right)=1.9 \mathrm{~m} / \mathrm{s}^2$

speed of transverse wave when lift is ging up $\left(v_1^2\right)=88 \mathrm{~m} / \mathrm{s}^2$

Using the formula for velocity

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

where $T$ is tension in string $\mu$ is mass/length.

$T=m a_{\mathrm{net}} \quad$ [mass of string]

Case I When lift goes up

$ \begin{aligned} a_{\mathrm{net}}=a_1+g & =10+2.1=12.1 \mathrm{~m} / \mathrm{s}^2 \\ T_1 & =m(12.1) \end{aligned} $

Case II When lift is going down

$ \begin{aligned} a_{\mathrm{nct}} & =g-a_2=10-1.9=8.1 \mathrm{~m} / \mathrm{s}^2 \\ T_2 & =m(8.1) \end{aligned} $

putting the value of $T_1, T_2$ in equation for $v$.

$ \begin{aligned} & v_1=\sqrt{\frac{T_1}{\mu}} \Rightarrow v_1=\sqrt{\frac{(12.1) m}{\mu}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & v_2=\sqrt{\frac{T_2}{\mu}} \Rightarrow v_2=\sqrt{\frac{(8.1) m}{\mu}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eq. (i) and (ii)

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

[ $\mu$ is same for both as mass]

$ \Rightarrow \quad v_2=v_1 \times \sqrt{\frac{T_2}{T_1}} \Rightarrow 72 \mathrm{~m} / \mathrm{s} $

Hence, the velocity is $72 \mathrm{~m} / \mathrm{s}$.