Waves

$f_{\text {beat }}=\left|f_1-f_2\right|$

and $f_{\text {open }}=2 \times f_{\text {closed }}$

When tubes are vibrated with both of their ends open, then

$ \begin{aligned} & f_{\text {beat }}^{\prime}=2\left|f_1-f_2\right| \\ & f'_{\text {beat }}=2 \times 2 \\ & f_{\text {beat }}^{\prime}=4 \end{aligned} $

Frequency of echo heard by the driver is given by,

$ \begin{aligned} f^{\prime} & =\frac{v+u}{v-u} f \\ f^{\prime} & =\frac{v+u}{v-u} \times n \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Difference in frequencies is $0.1 n$

$ \begin{array}{ll} \therefore & f^{\prime}-n=0.1 n \\ & f^{\prime}=1.1 n \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i) and (ii)

$ \begin{aligned} & \frac{v+u}{v-u}=1.1 \\ & \frac{336+u}{336-u}=1.1 \\ & 2.1 u=33.6 \\ & u=16 \mathrm{~ms}^{-1} \end{aligned} $

The tube is 50 cm long and is closed at one end.

Finding the Wavelength:

For the fifth harmonic in a closed pipe, the formula is:

$ \frac{5 \lambda}{4} = L $

This means five-fourths of the wavelength fits into the total length ($L$) of the tube.

Let's solve for wavelength ($\lambda$):

$ \lambda = \frac{4L}{5} = \frac{4 \times 50}{5} = 40~\mathrm{cm} $

Calculating the Phase Difference:

The phase difference ($\Delta \phi$) between two points separated by a distance $x$ is:

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

Here, $x = 42 ~\mathrm{cm}$ (the distance from the open end to the other particle) and $\lambda = 40~\mathrm{cm}$:

$ \Delta \phi = \frac{2\pi}{40} \times 42 $

This can be calculated as:

$ \Delta \phi = \frac{2\pi \times 42}{40} = \frac{84\pi}{40} = \frac{21\pi}{10} $

If you convert this to degrees:

$ \Delta \phi = \frac{21\pi}{10} \times \frac{180^\circ}{\pi} = 378^\circ $

But phases repeat every 360°. So, we subtract 360° to get the simpler value:

$ 378^\circ - 360^\circ = 18^\circ $

So, the phase difference is $18^\circ$.

For fundamental frequency, at points $A$ and $B$, there should be an antinode and at point $C$, there should be a node.

$ \begin{aligned} \Rightarrow & & \frac{\lambda}{4}+\frac{\lambda}{4} & =L \\ \Rightarrow & & \lambda & =2 L=2 \times 125=250 \mathrm{~cm} \end{aligned} $

For frequency,

$ \begin{aligned} v & =\frac{v}{\lambda}=\frac{5000}{250 \times 10^{-2}} \\ & =2000 \mathrm{~Hz} \text { or } 2 \mathrm{kHz} \end{aligned} $

Beat frequency

$ f=|323-320|=3 \mathrm{~Hz} $

∴ Beat period

$ T=\frac{1}{f}=\frac{1}{3} \mathrm{~s} $

The time interval between a maximum sound and its adjacent minimum sound is equal to half of the beat period.

i.e., Time interval $=\frac{T}{2}=\frac{\frac{1}{3}}{2}=\frac{1}{6} \mathrm{~s}$

Step 1: Understanding the Doppler Effect for a Moving Observer

When the sound source is not moving and the observer moves toward it, the frequency they hear increases.

When the observer moves away from the stationary source, the frequency they hear decreases.

Step 2: Writing the Frequency Formulas

If the actual frequency of the source is $n$, the speed of sound is $v$, and the observer's speed is $u$:

When moving toward the source: $ n_1 = n \left( \frac{v + u}{v} \right) $

When moving away from the source: $ n_2 = n \left( \frac{v - u}{v} \right) $

Step 3: Using the Ratio Given in the Problem

We are told that the ratio of the two frequencies is $n_1 : n_2 = 71 : 65$. We set up an equation:

$ \frac{n_1}{n_2} = \frac{v + u}{v - u} = \frac{71}{65} $

Step 4: Solving for Observer's Speed $(u)$

Cross-multiply the equation: $ 65(v + u) = 71(v - u) $

Expand: $ 65v + 65u = 71v - 71u $

Move both $v$ terms to one side and $u$ terms to the other: $ 65u + 71u = 71v - 65v $ $ 136u = 6v $

Divide both sides by 136: $ u = \frac{6v}{136} = \frac{3v}{68} $

Step 5: Substitute Speed of Sound Value

The speed of sound is $v = 340~\mathrm{m/s}$.

Plug this value into the formula for $u$: $ u = \frac{3 \times 340}{68} = \frac{1020}{68} = 15~\mathrm{m/s} $

To change $15~\mathrm{m/s}$ into $\mathrm{km/h}$, multiply by $\frac{18}{5}$: $ 15 \times \frac{18}{5} = 54~\mathrm{km/h} $

Step 1: Find the amplitude.

The total distance from one extreme point to the other is 10 cm. So the amplitude is half of this: $5~\mathrm{cm} = 0.05~\mathrm{m}$.

Step 2: Find the angular frequency ($( \omega )$).

The frequency $f$ is $210~\mathrm{Hz}$.

Angular frequency: $\omega = 2\pi f = 2 \times \frac{22}{7} \times 210 = 44 \times 30 = 1320~\mathrm{rad/s}$.

Step 3: Find the wave number ($k$).

The speed $v$ is $330~\mathrm{m/s}$.

Wave number: $k = \frac{\omega}{v} = \frac{1320}{330} = 4~\mathrm{rad/m}$.

Step 4: Write the equation of the wave.

The standard equation for a wave traveling along the positive $X$-axis is $s(x, t) = A \sin(kx - \omega t)$.

Here, $A = 0.05~\mathrm{m}$, $k = 4~\mathrm{rad/m}$, and $\omega = 1320~\mathrm{rad/s}$, so:

$ \begin{aligned} s(x, t) &= 0.05 \sin(4x - 1320t) \end{aligned} $

Step 1: Formula for Fundamental Frequency

The basic formula for the frequency of a vibrating string is:

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

Here, $f$ is frequency, $l$ is the length of the string, $T$ is the tension, and $\mu$ is mass per unit length (which stays the same in this problem).

Step 2: Frequency with Initial Length and Tension

The first frequency, with length $l$ and tension $T_1$ is:

$ f_1 = \frac{1}{2l} \sqrt{\frac{T_1}{\mu}} \tag{i} $

Step 3: Frequency After Changing Length and Tension

The length of the string is shortened by $25\%$, so the new length is $75\%$ of $l$, or $\frac{3}{4}l$. The new tension is $T_2$ and the new frequency becomes $2f_1$ (it increases by $100\%$):

$ 2f_1 = \frac{1}{2(\frac{3}{4}l)} \sqrt{\frac{T_2}{\mu}} \tag{ii} $

Step 4: Set Up the Equation

Now, make it easier by expressing $2f_1$ in terms of the formula:

$ 2f_1 = \frac{1}{(3/2)l} \sqrt{\frac{T_2}{\mu}} $

Because splitting the denominator gives $2 \times \frac{1}{2(\frac{3}{4}l)} = \frac{1}{(3/2)l}$.

Step 5: Plug $f_1$ from Step 2 into the New Equation

Replace $f_1$ in the new formula using the old frequency formula:

$ 2 \left( \frac{1}{2l} \sqrt{\frac{T_1}{\mu}} \right) = \frac{1}{(3/2)l} \sqrt{\frac{T_2}{\mu}} $

Step 6: Simplify the Equation

Multiply both sides to clear out the denominators:

$ \frac{1}{l} \sqrt{\frac{T_1}{\mu}} = \frac{2}{3l} \sqrt{\frac{T_2}{\mu}} $

Multiply both sides by $3l$:

$ 3\sqrt{\frac{T_1}{\mu}} = 2\sqrt{\frac{T_2}{\mu}} $

Step 7: Get Rid of Square Roots by Squaring Both Sides

Square each side to eliminate the square roots:

$ 9\frac{T_1}{\mu} = 4\frac{T_2}{\mu} $

The $\mu$ cancels out:

$ 9T_1 = 4T_2 $

Step 8: Find $\frac{T_2}{T_1}$

Divide both sides by $4T_1$ to solve for $\frac{T_2}{T_1}$:

$ \frac{T_2}{T_1} = \frac{9}{4} $

Step 1: Write the formula for frequency in an open pipe

The frequency of the $n$th harmonic in an open organ pipe is:

$ f_0=\frac{n v}{2 l_0} $

Step 2: Find the frequency for the 3rd harmonic in the open pipe

For the third harmonic, $n = 3$:

$ f_0' = \frac{3 v}{2 l_0} $

Step 3: Write the formula for frequency in a closed pipe

The frequency of the $n$th harmonic in a closed pipe is:

$ f_c=\frac{n v}{4 l_c} $

Step 4: Find the frequency for the 5th harmonic in the closed pipe

For the fifth harmonic, $n = 5$:

$ f_c' = \frac{5 v}{4 l_c} $

Step 5: Write the ratio of the two frequencies

We want to find the ratio $\frac{f_0'}{f_c'}$.

$ \frac{f_0'}{f_c'} = \frac{\frac{3v}{2l_0}}{\frac{5v}{4l_c}} $

Step 6: Simplify the ratio

First, simplify by dividing:

$ \frac{3v}{2l_0} \times \frac{4l_c}{5v} = \frac{12l_c}{10l_0} $

Step 7: Use the ratio of lengths

It is given that the ratio of lengths of the open and closed pipes is $2:3$, so $\frac{l_0}{l_c} = \frac{2}{3}$.

Step 8: Substitute the value for $\frac{l_0}{l_c}$

$ \frac{12}{10} \times \frac{1}{\frac{l_0}{l_c}} = \frac{12}{10} \times \frac{1}{\frac{2}{3}} = \frac{12}{10} \times \frac{3}{2} = \frac{36}{20} = \frac{9}{5} $

Step 9: State the final ratio

The ratio of the frequencies is $9:5$.

The standard form of a travelling wave is

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

Comparing

$ \begin{aligned} & y=3 \sin (4 x+200 t) \\ & k=4 \mathrm{rad} / \mathrm{m} \\ & \omega=\mathrm{rad} / \mathrm{sec} \end{aligned} $

wave speed is given by

$ v=\frac{\omega}{k}=\frac{200}{4}=50 \mathrm{~m} / \mathrm{s} $

Wave speed for string

$ \begin{aligned} & v=\sqrt{\frac{T}{\mu}} \Rightarrow \mu=\frac{T}{v^2} \\ & \mu=\frac{500}{(50)^2}=\frac{500}{2500}=0.2 \mathrm{~kg} / \mathrm{m} \end{aligned} $

The initial fundamental frequency

$ \begin{aligned} & f_1=\frac{1}{2 L_1} \sqrt{\frac{T_1}{\mu}} \\ & 300=\frac{1}{2 L_1} \sqrt{\frac{T_1}{\mu}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

The new fundamental frequency

$ \begin{aligned} & f_2=\frac{1}{2 L_2} \sqrt{\frac{T_2}{\mu}} \\ & 100=\frac{1}{2\left(2 L_1\right)} \sqrt{\frac{T_2}{\mu}} \\ & 100=\frac{1}{4 L_1} \sqrt{\frac{T_2}{\mu}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Divide the Eq. (i) by the Eq. (ii)

$ \begin{aligned} \frac{300}{100} & =\frac{\frac{1}{2 L_1} \sqrt{\frac{T_1}{\mu}}}{\frac{1}{4 L_1} \sqrt{\frac{T_2}{\mu}}} \\ 3 & =\frac{\frac{1}{2} \sqrt{T_1}}{\frac{1}{4} \sqrt{T_2}} \Rightarrow 3=\frac{4}{2} \sqrt{\frac{T_1}{T_2}} \\ 3 & =2 \sqrt{\frac{T_1}{T_2}}=\frac{3}{2}=\sqrt{\frac{T_1}{T_2}}=\frac{9}{4}=\frac{T_1}{T_2} \\ \frac{T_2}{T_1} & =\frac{4}{9} \end{aligned} $

The ratio of the new tension to the initial tension is $\frac{4}{9}$.

$I_R=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi$

Here, $\phi=\frac{\pi}{2}$ and $I_1=I_2=I$

$ \therefore \quad I_R=I+I+2 \sqrt{I \cdot I} \cos \frac{\pi}{2}=2 I $

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} $

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 $72 \mathrm{~m} / \mathrm{s}$.

Transverse waves are those waves whose vibration is perpendicular to direction of propagation.

It can propagate through solid and liquid. Longitudinal waves are those wave's whose vibration of medium is parallel to direction of propagation.

It can propagate in all mediums (solid, liquid or gases).

No mechanical waves can propagate without any medium.

So, option (c) is the correct option.

Given,

Observer velocity, $v_O=10 \mathrm{~m} / \mathrm{s}$

Source velocity, $v_S=-10 \mathrm{~m} / \mathrm{s}$

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

Source frequency $f^{\prime}=1980 \mathrm{~Hz}$

According to Doppler's effect formula

$ \begin{aligned} f^{\prime} & =f\left(\frac{v-v_0}{v-v_s}\right) \\ 1980 & =f\left(\frac{340-10}{340+10}\right) \\ f & =\frac{1980 \times 350}{330} \\ f & =2100 \mathrm{~Hz} \end{aligned} $

Given, displacement

$ y=5 \times 10^3 \sin \left(125 \pi x-\frac{\pi}{2} t\right) $

comparing it with standard wave equation,

$ \begin{aligned} & y=A \sin (\omega t+k x), \text { we get } \\ & A=5 \times 10^{-3} \mathrm{~m} \\ & K=125 \pi \\ & \omega=\frac{\pi}{2} \end{aligned} $

We know that, $K=\frac{2 \pi}{\lambda}$

Thus, $\lambda=\frac{2 \pi}{K}=\frac{2 \pi}{125 \pi}=0.16 \mathrm{~m}$

Also $T=\frac{2 \pi}{\omega}$

Thus, $T=\frac{2 \pi}{\pi / 2}=4 \mathrm{~s}$

Given, $v_A=250 \mathrm{~Hz}$

Beats $=5$ beats $/ \mathrm{s}$

$ \begin{array}{ll} & v_2-v_1=5 \\ \Rightarrow & v_2=5+250=255 \mathrm{~Hz} \end{array} $

For closed organ pipe $v_1=\frac{v}{4 L_1}$

For 7 th harmonic; $7 v_1=\frac{7 v}{4 L_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For an open pipe, fundamental energy is $v=\frac{v}{2 L_2}$

For 7th harmonic of open pipe

$ 4 v_2=\frac{4 v}{2 L_2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} &\text { Therefore from Eqs. (i), and (ii) we have }\\ &\begin{aligned} 7 v_1 & =4 v_2 \\ \Rightarrow \quad \frac{7 v}{4 L_1} & =\frac{4 v}{2 L_2} \\ & =\frac{L_1}{L_2}=\frac{14}{16}=\frac{7}{8}=7: 8 \end{aligned} \end{aligned} $

Apparent frequency is given by

$ f=f_0\left(\frac{v_s+v_0}{v_S}\right)=f_0\left(\frac{v+\frac{v}{S}}{5}\right)=\frac{6}{5} f_0 $

Hence, percentage increase in frequency

$ =\frac{\left(\frac{6}{5} f_0-f_0\right)}{f_0} \times 100 \%=20 \% $

Given,

Length of $\operatorname{rod} B D=L$

Length of string $A B=C D=1$

Mass suspended $=M$

Length of $B O=x$

$ \text { Length of } O D=L-x $

Frequency of 1st harmonics of $A B=$ Frequency of 2nd harmonics of $C D$. Let $T_{A B}$ is tension in string $A B . T_{C D}$ is the tension in string $C D$.

Frequency of 1st harmonics of $A B$ is given by

$ f=\frac{1}{2 l} \sqrt{\frac{T_{A B}}{m}}, l=\text { length of string }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Frequency of 2nd harmonics of $C D$

$ f=\frac{2}{2} \sqrt{\frac{T_{C D}}{m}}=\frac{1}{l} \sqrt{\frac{T_{C D}}{m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

Eq. (i) $=$ Eq. (ii) (given)

$ \begin{aligned} \Rightarrow & \frac{1}{2 l} \sqrt{\frac{T_{A B}}{m}} =\frac{1}{l} \sqrt{\frac{T_{C D}}{m}} \\ \Rightarrow & T_{A B} =4 T_{C D}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

For rotational equilibrium of massless rod, taking torque at point $O$.

$ \begin{aligned} & \tau=r \times F(\text { Here } F=\text { Tension }) \\ & T_{A B} \times(X)=T_{C D}(L-X) \end{aligned} $

Put value of $T_{A B}$, then

$ X=L / 5 $

When an observer moves towards a stationary sound source, the apparent frequency they perceive increases. This is due to the observer's motion, which causes them to encounter sound waves more frequently than if they were stationary.

In this scenario, the observer's speed is $\frac{1}{5}$ of the speed of sound. The key parameters are:

Observer's speed, $v_0 = \frac{v}{5}$

Source speed, $v_s = 0$ (stationary)

Frequency of the sound emitted by the source, $f$

Speed of sound, $v$

The formula to calculate the apparent frequency ($f'$) when the observer is moving towards a stationary source is:

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

Substituting the given values:

$ f' = \left(\frac{v + \frac{v}{5}}{v}\right) f $

Simplifying the equation:

$ f' = \left(\frac{6v}{5v}\right) f = \frac{6}{5}f = 1.2f $

Therefore, the apparent frequency recorded by the observer is $1.2f$.

Speed of wave on the string is given by

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

where $T$ is tension, $\mu$ is linear mass density.Mass of the rope $=m$

$ \mu=\frac{m}{L} $

, iii)

Put the value of $T$ and $\mu$ in Eq. (i),

$ v=\sqrt{\frac{m g}{\frac{m}{L}}}=\sqrt{g L} $

Now, using equation of motion,

$ v^2=u^2+2 a s $

( $v=0$, rope is connected to ceiling)

$ \begin{aligned} & a=\frac{U^2}{2 S} \Rightarrow \frac{g L}{2 L} \\ & \quad \quad \quad \text { (because } S=L \text { ) } \\ & a=\frac{g}{2} \end{aligned} $

Hence, the value of acceleration is constant.

Let's analyze the given wave equation. The equation is:

$ y = (0.02) \sin (\pi x - 8\pi t) $

Here, the wave is in the standard form:

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

where:

$A = 0.02$ is the amplitude.

$k$ is the wave number.

$\omega$ is the angular frequency.

From the given equation:

The wave number is: $k = \pi$.

The angular frequency is: $\omega = 8\pi$.

The wave velocity $v$ (also called the phase velocity) is given by the formula:

$ v = \frac{\omega}{k} $

Substitute the known values:

$ v = \frac{8\pi}{\pi} = 8 \, \text{ms}^{-1} $

Thus, the velocity of the wave is $8 \, \text{ms}^{-1}$, which corresponds to Option C.

We know that, fundamental frequency produced in cylindrical open tube

$ f=\frac{V}{2 L} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where $L=$ length of tube When tube is dipped vertically in water so that half of its portion is in water, then it acts a closed pipe of length half of original length $\left(\frac{L}{2}\right)$.

So, fundamental frequency of this closed tube,

$ f^{\prime}=\frac{v}{4\left(\frac{L}{2}\right)}=\frac{v}{2 L}=f \,\,\,\,\,[from Eq. (i)]$

Maximum speed of wave is given as

$ \begin{aligned} v_{\max } & =\omega A \\ & =\text { Angular speed } \text { × } \text { Amplitude } \end{aligned} $

Thus,

In option (a) $y(x, t)=2 \sin (2 x-2 t)$

$ \begin{aligned} & & A & =2, \omega=2 \\ & \therefore & v_{\text {max }} & =2 \times 2=4 \mathrm{~m} / \mathrm{s} \end{aligned} $

In option (b), $y(x, t)=3 \sin (2 x-3 t)$

$ \begin{array}{rlrl} & & A & =3, \omega=3 \\ \therefore & & v_{\max } & =3 \times 3 =9 \mathrm{~m} / \mathrm{s} \end{array} $

In option (c), $y(x, t)=2 \sin (3 x-2 t)$

$ \begin{aligned} A & =2, \omega=2 \\ \therefore \quad v_{\max } & =2 \times 2=4 \mathrm{~m} / \mathrm{s} \end{aligned} $

In option (d), $y(x, t)=3 \sin (5 x-2 t)$

$ \begin{aligned} A & =3, \omega=2 \\ \therefore \quad v_{\max } & =3 \times 2=6 \mathrm{~m} / \mathrm{s} \end{aligned} $

Clearly, wave speed is mximum as 9 $\mathrm{m} / \mathrm{s}$ in wave equation given in option (b).

Initial length of wire, $I_1=0.4 \mathrm{~m}$

Frequency of stretched wire,

$ f_1=250 \mathrm{vib} / \mathrm{s}=250 \mathrm{~Hz} $

Final length of wire after stretching.

$ \begin{aligned} & l_2=l_1+0.1 \mathrm{~m} \\ & =0.4+0.1=0.5 \mathrm{~m} \end{aligned} $

Initial force (tension), $T_1=T$

Final force (tension), $T_2=\frac{T}{4}$

We know that, frequency of wave produced due to stretched string.

$ \begin{aligned} f & =\frac{l}{2} \sqrt{\frac{T}{m}} \\ \Rightarrow \quad \frac{f_2}{f_1} & =\frac{l_1}{l_2} \cdot \sqrt{\frac{T_2}{T_1}}=\frac{0.4}{0.5} \sqrt{\frac{T / 4}{T}} \\ & =\frac{4}{5} \times \frac{1}{2} \\ \frac{f_2}{f_1} & =\frac{2}{5} \\ \Rightarrow \quad f_2 & =\frac{2}{5} f_1=\frac{2}{5} \times 250=100 \mathrm{~Hz} \end{aligned} $

Beat frequency produced by two strings $A$ and $B=\Delta f_1$

When tension in string $A$ is slightly increased, then beat of frequency $=\Delta f_2$ Since, on increasing tension in the string $A$, beat frequency $\Delta f_2$ decreases from initial beat frequency $\Delta f_1$. Hence, frequency of string $B$ must be greater than fundamental frequency of string A. i.e, frequency of string $B=f_0+\Delta f_1$

The distance between two successive minima is equal to wavelength of transverse wave.

l.e. $\lambda=2.7 \mathrm{~m}$

Distance travelled by wave in 15 s is equal to $5 \lambda$ as five crests of the wave pass a given point.

i.e $d=\lambda=5 \times 2.7=13.5 \mathrm{~m}$

and $t=15 \mathrm{~s}$

∴ Speed of wave, $v=\frac{d}{t}=\frac{13.5}{1.5}=0.9 \mathrm{~m} / \mathrm{s}$

The ratio between the maximum and minimum intensities of the resultant wave,

$ \begin{aligned} & I_{\max }: I_{\min }=9: 4 \\ & \Rightarrow \frac{I_{\max }}{I_{\min }}=\frac{9}{4} \Rightarrow \frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2}=\left(\frac{3}{2}\right)^2 \\ & \Rightarrow \frac{A_1^2\left(1+\frac{A_2}{A_1}\right)^2}{A_1^2\left(1-\frac{A_2}{A_1}\right)^2}=\left(\frac{3}{2}\right)^2 \\ & \Rightarrow \frac{\left(1+\frac{A_2}{A_1}\right)^2}{\left(1-\frac{A_2}{A_1}\right)^2}=\left(\frac{3}{2}\right)^2 \Rightarrow \frac{1+\frac{A_2}{A_1}}{1-\frac{A_2}{A_1}}=\frac{3}{2} \\ & \Rightarrow \frac{A_1+A_2}{A_1-A_2}=\frac{3}{2} \\ & \Rightarrow 2 A_1+2 A_2=3 A_1-3 A_2 \\ & \Rightarrow 5 A_2=A_1 \Rightarrow \frac{A_2}{A_1}=\frac{1}{5}=0.20 \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} $