Waves

Before waxing the fork, A

The beat frequency is the number of beats per second.

$ \mathrm{f}_{\mathrm{b}}=\frac{\text { Number of beats }}{\text { Time }}=\frac{8 \text { beats }}{2 \mathrm{~s}}=4 \mathrm{~Hz} $

This means the absolute difference between the frequencies of fork $A\left(f_A\right)$ and fork $B\left(f_B\right)$ is 4 Hz :

$ \left|\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}\right|=4 \mathrm{~Hz} $

Since $\mathrm{f}_{\mathrm{B}}=380 \mathrm{~Hz}$, the original frequency of fork A could be either :

Case 1 : $\mathrm{f}_{\mathrm{A}}=380+4=384 \mathrm{~Hz}$

Case 2 : $\mathrm{f}_{\mathrm{A}}=380-4=376 \mathrm{~Hz}$

When a tuning fork is loaded with wax, its mass increases, which causes its frequency to decrease. Let the new frequency of fork $A$ be $f_A^{\prime}$, where $f_A^{\prime} < f_A$.

The new beat frequency ( $\mathrm{f}_{\mathrm{b}}^{\prime}$ ) is :

$ \mathrm{f}_{\mathrm{b}}^{\prime}=\frac{4 \text { beats }}{2 \mathrm{~s}}=2 \mathrm{~Hz} $

Initially, $\mathrm{f}_{\mathrm{A}}$ is higher than $\mathrm{f}_{\mathrm{B}}$ by $4 \mathrm{~Hz}(384-380=4)$.

If $f_A$ decreases (to say 382 Hz ), the gap between $f_A$ and $f_B$ decreases $(382-380=2)$.

The original frequency of tuning fork A must have been higher than fork B to allow the beat frequency to decrease as $f_A$ dropped.

Therefore, the original frequency of tuning fork A is 384 Hz . Hence, the correct answer is $\mathbf{3 8 4}$.

The velocity of sound in an ideal gas depends on the absolute temperature of the medium.

The speed of sound (v) in air is given by the formula :

$ \mathrm{v}=\sqrt{\frac{\gamma \mathrm{RT}}{\mathrm{M}}} $

From this, we can see that :

$ \mathrm{v} \propto \sqrt{\mathrm{~T}} $

where T is the absolute temperature in Kelvin $(\mathrm{K})$.

Initial State :

Temperature $=\mathrm{T}_1=0^{\circ} \mathrm{C}=(0+273) \mathrm{K}=273 \mathrm{~K}$

Velocity $=\mathrm{v}_1$

Final State :

Temperature $=\mathrm{T}_2=\alpha^{\circ} \mathrm{C}=(\alpha+273) \mathrm{K}$

Velocity $=\mathrm{v}_2=2 \mathrm{v}_1($ since the velocity is doubled $)$.

Using the proportionality $\mathrm{v} \propto \sqrt{\mathrm{T}}$ :

$ \frac{\mathrm{v}_2}{\mathrm{v}_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} $

Substituting the values :

$ \frac{2 v_1}{v_1}=\sqrt{\frac{\alpha+273}{273}} $

$\Rightarrow $ $4=\frac{\alpha+273}{273}$

$\Rightarrow $ $4 \times 273=\alpha+273$

$\Rightarrow $ $\alpha=1092-273$

$\Rightarrow $ $ \alpha=819 $

Therefore, the value of $\alpha$ is 819 . Hence, the correct answer is $\mathbf{8 1 9}$.

When sound waves from two coherent sources (like the two loudspeakers) meet, they interfere with each other.

Constructive Interference (Maxima) occurs when the waves are in phase.

The path difference, $\Delta \mathrm{x}$ is an integer multiple of the wavelength ( $\lambda$ ).

$ \Delta \mathrm{x}=\mathrm{n} \lambda \ldots(\text { where } \mathrm{n}=0,1,2, \ldots) $

Destructive Interference (Minima) occurs when the waves are out of phase. This happens when the path difference is a half-integer multiple of the wavelength.

$ \Delta \mathrm{x}=\left(\mathrm{n}+\frac{1}{2}\right) \lambda $

It is given that A is equidistant from both speakers $\left(\mathrm{L}_{1 \mathrm{~A}}=\mathrm{L}_{2 \mathrm{~A}}\right)$. Therefore, the path difference at A is zero $\left(\Delta \mathrm{x}_{\mathrm{A}}=0\right)$. This corresponds to the central maximum ( $\mathrm{n}=0$ ).

As the recorder moves towards $B$, the distance to $L_2$ increases more than the distance to $L_1$, so the path difference ( $\Delta \mathrm{x}$ ) steadily increases.

The signal undergoes 10 cycles of minima and maxima. One complete cycle means moving from one maximum to the next consecutive maximum, which corresponds to an increase in path difference by exactly one wavelength ( $\lambda$ ).

Since it goes through 10 full cycles starting from the central maximum, the path difference at point B must be exactly 10 wavelengths.

$ \Delta \mathrm{x}_{\mathrm{B}}=10 \lambda $

The path length from $L_1$ to $B$ is,

$ \mathrm{x}_{\mathrm{L} 1}=\sqrt{40^2+(25-5)^2} $

$\Rightarrow $ $\mathrm{x}_{\mathrm{L} 1}=\sqrt{40^2+20^2}=\sqrt{2000}$

$\Rightarrow $ $ \mathrm{x}_{\mathrm{L} 1}=20 \sqrt{5} \mathrm{~m} $

The path length from $L_2$ to $B$ is,

$ \mathrm{x}_{\mathrm{L} 2}=\sqrt{(40)^2+(25+5)^2}=\sqrt{1600+900} $

$\Rightarrow $ $ \mathrm{x}_{\mathrm{L} 2}=\sqrt{2500}=50 \mathrm{~m} $

So, the path difference is,

$ \Delta \mathrm{x}_{\mathrm{B}}=\mathrm{x}_{\mathrm{L} 2}-\mathrm{x}_{\mathrm{L} 1} $

$\Rightarrow $ $ \Delta \mathrm{x}_{\mathrm{B}}=50-20 \sqrt{5} $

We are given $\sqrt{5}=2.23$.

$ \Delta \mathrm{x}_{\mathrm{B}}=50-20(2.23) $

$\Rightarrow $ $\Delta \mathrm{x}_{\mathrm{B}}=50-44.6$

$\Rightarrow $ $\Delta \mathrm{x}_{\mathrm{B}}=5.4 \mathrm{~m}$

$\Rightarrow 10 \lambda=\Delta \mathrm{x}_{\mathrm{B}}$

$ \Rightarrow \lambda=0.54 \mathrm{~m} $

Using the relation between wave speed (v), frequency (f), and wavelength ($\lambda$) :

$ \mathrm{v}=\mathrm{f} \cdot \lambda $

We are given the speed of sound $\mathrm{v}=324 \mathrm{~m} / \mathrm{s}$.

$ \mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{324}{0.54}=600 \mathrm{~Hz} $

Therefore, the frequency of the input signal is 600.

Step 1: Write the formulas for the frequencies of open and closed organ pipes.

The frequency of an open organ pipe is $f_o = \frac{v}{2l}$, where v is the speed of sound and l is the length of the pipe.

The frequency of a closed organ pipe is $f_c = \frac{v}{4l}$.

Step 2: Find the frequency of the seventh overtone for each pipe.

The seventh overtone in an open pipe is the 8th harmonic (since the first overtone is the 2nd harmonic). So, its frequency is: $f_{o_7} = 8 \times \frac{v}{2l}$.

In a closed pipe, only odd harmonics are present. The seventh overtone is the 15th harmonic, so: $f_{c_7} = 15 \times \frac{v}{4l}$.

Step 3: Find the ratio of the seventh overtone frequencies.

The ratio of the frequencies is: $\frac{f_{c_7}}{f_{o_7}} = \frac{15 \frac{v}{4l}}{8 \frac{v}{2l}}$.

Simplify the ratio: $\frac{15}{4l} \div \frac{8}{2l} = \frac{15}{4l} \times \frac{2l}{8} = \frac{30}{32} = \frac{15}{16}$.

Step 4: Connect the answer to the value of a

The given ratio is $\frac{a-1}{a}$, so: $\frac{a-1}{a} = \frac{15}{16}$.

Solve for $a$: $a = 16$.

To solve this problem, let's first understand how the harmonics of open organ pipes work. For an open organ pipe, the harmonics are given by the formula:

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

where:

• $f_n$ is the frequency of the nth harmonic,


• $n$ is the harmonic number (an integer),


• $v$ is the speed of sound in air,


• $L$ is the length of the organ pipe.

Given that the speeds of sound in air $v = 333 \, \text{m/s}$, and the lengths of the two open organ pipes are $60 \, \text{cm} = 0.60 \, \text{m}$ and $90 \, \text{cm} = 0.90 \, \text{m}$, we can calculate the frequencies of the 6th harmonic for the 60 cm pipe and the 5th harmonic for the 90 cm pipe.

For the 60 cm pipe at the 6th harmonic ($n = 6$):

$f_{6,60} = 6 \frac{333}{2 \times 0.60} = 6 \times \frac{333}{1.2} = 6 \times 277.5 = 1665 \, \text{Hz}$

For the 90 cm pipe at the 5th harmonic ($n = 5$):

$f_{5,90} = 5 \frac{333}{2 \times 0.90} = 5 \times \frac{333}{1.8} = 5 \times 185 = 925 \, \text{Hz}$

The difference in frequencies between these two modes is:

$\Delta f = f_{6,60} - f_{5,90} = 1665 \, \text{Hz} - 925 \, \text{Hz} = 740 \, \text{Hz}$

Therefore, the difference of frequencies for the given modes is $740 \, \text{Hz}$.

The fundamental frequency of a string (in this case, a sonometer wire) when it is vibrating, is inversely proportional to its length, provided the tension in the string and the linear mass density (mass per unit length) remain constant. This relationship is given by the formula:

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

where:

• $f$ is the frequency of the string,
• $L$ is the length of the string,
• $T$ is the tension in the string, and
• $\mu$ is the linear mass density of the string.

Given that the sonometer wire has a resonating length of 90 cm (0.9 m) with a fundamental frequency of $400 \mathrm{Hz}$, we can set up our first equation but note that we are comparing two states of the same string under the same tension and hence can eliminate the tension and density terms for comparison purposes:

$400 = \frac{1}{2 \times 0.9} \sqrt{\frac{T}{\mu}}$

For the second scenario where the fundamental frequency is $600 \mathrm{Hz}$, we are asked to find the new length $L_2$. We can set up the equation in a similar manner:

$600 = \frac{1}{2L_2} \sqrt{\frac{T}{\mu}}$

Since the tension $T$ and the linear density $\mu$ are constants, and they do not change between the two states, we can set up a proportion between the two states by dividing the second equation by the first, which yields:

$\frac{600}{400} = \frac{\frac{1}{2L_2}}{\frac{1}{2 \times 0.9}}$

Simplifying this equation gives:

$\frac{600}{400} = \frac{0.9}{L_2}$

or

$\frac{3}{2} = \frac{0.9}{L_2}$

Solving for $L_2$ gives:

$L_2 = \frac{0.9 \times 2}{3}$

Calculating the value:

$L_2 = \frac{1.8}{3} = 0.6 \hspace{1mm} \text{meters}$

Converting meters to centimeters (since 1 meter = 100 centimeters), we find:

$L_2 = 0.6 \times 100 = 60 \hspace{1mm} \text{centimeters}$

Therefore, the resonating length of the wire with a fundamental frequency of $600 \mathrm{Hz}$ under the same tension is 60 cm.

To solve this problem, we'll have to use the relationship between the frequency of a vibrating string and the tension applied to it. When a tuning fork resonates with a sonometer wire, their frequencies are equal.

The frequency of a vibrating string is given by the formula:

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

Where:

• $ f $ is the frequency of the vibration


• $ L $ is the length of the wire


• $ T $ is the tension in the wire


• $ \mu $ is the linear mass density of the wire

The linear mass density ($ \mu $) of the wire remains constant.

Initially, when the string resonates with the tuning fork, the frequency of both is given by:

$f_1 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}}$

Here $ T_1 = 6 \mathrm{~N} $ and $ L = 1 \mathrm{~m} $, so we have:

$f_1 = \frac{1}{2 \cdot 1} \sqrt{\frac{6}{\mu}} = \frac{1}{2} \sqrt{\frac{6}{\mu}}$

Now, when the tension is changed to $ T_2 = 54 \mathrm{~N} $, the frequency of the wire changes to $ f_2 $ and it is given by:

$f_2 = \frac{1}{2L} \sqrt{\frac{T_2}{\mu}}$

Since $ L $ and $ \mu $ remain the same, substituting $ T_2 $:

$f_2 = \frac{1}{2 \cdot 1} \sqrt{\frac{54}{\mu}} = \frac{1}{2} \sqrt{\frac{54}{\mu}}$

Notice that $ 54 = 6 \times 9 $, therefore:

$f_2 = \frac{1}{2} \sqrt{\frac{6 \times 9}{\mu}} = \frac{1}{2} \sqrt{9} \sqrt{\frac{6}{\mu}} = \frac{3}{2} \sqrt{\frac{6}{\mu}} = 3 f_1$

When the tension was increased, the frequency became thrice the original frequency.

Since the second instance of the string produces 12 beats per second this means that the frequency of the tuning fork (and original string) and the new frequency (of the string with higher tension) differ by 12 Hz. If $ f_F $ is the frequency of the tuning fork, then:

Either $ f_2 = f_F + 12 $ Hz or $ f_2 = f_F - 12 $ Hz.

Since we have determined $ f_2 = 3 f_1 $ and $ f_1 = f_F $, we can state:

Either $ 3f_F = f_F + 12 $ or $ 3f_F = f_F - 12 $.

If $ 3f_F = f_F - 12 $, then :

$3f_F - f_F = -12$

$2f_F = -12$

This result is not possible since frequency cannot be negative.

So, we must consider the correct equation which is:

$f_2 = f_F + 12$

Now, substituting $ f_2 = 3f_F $:

$3f_F = f_F + 12$

$2f_F = 12$

$f_F = 6 \text{ Hz}$

Thus, the frequency of the tuning fork is 6 Hz.

To solve this problem, we need to understand the relationship between the intensity of sound waves and the distance from the source. The intensity $I$ of sound waves from a point source decreases with the square of the distance $r$ from the source, according to the inverse square law, which can be expressed as:

$ I \propto \frac{1}{r^2} $

This means that if the distance is doubled, the intensity becomes one-fourth of its initial value because $ (2r)^2 = 4r^2 $.

The initial intensity given at the origin (source) is:

$ I_0 = 16 \times 10^{-8} \mathrm{~Wm}^{-2} $

Let's call $ I_1 $ the intensity at $ r = 2 \text{ m} $ and $ I_2 $ the intensity at $ r = 4 \text{ m} $. Using the inverse square law, we can write:

$ I_1 = \frac{I_0}{(2)^2} = \frac{I_0}{4} $

and

$ I_2 = \frac{I_0}{(4)^2} = \frac{I_0}{16} $

Now substitute the given value for $ I_0 $ to find $ I_1 $ and $ I_2 $:

$ I_1 = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8} \mathrm{~Wm}^{-2} $

$ I_2 = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8} \mathrm{~Wm}^{-2} $

Now to find the difference in intensity (magnitude only) between the two points, we subtract $ I_2 $ from $ I_1 $:

$ \Delta I = | I_1 - I_2 | $

$ \Delta I = | 4 \times 10^{-8} - 1 \times 10^{-8} | $

$ \Delta I = 3 \times 10^{-8} \mathrm{~Wm}^{-2} $

So the difference in intensity (magnitude only) at the two points is:

$ 3 \times 10^{-8} \mathrm{~Wm}^{-2} $

$\begin{aligned} & \frac{V}{4 \ell_1}=30 \Rightarrow \ell_1=\frac{11}{4} m \\ & \frac{V}{4 \ell_2}=110 \Rightarrow \ell_2=\frac{3}{4} m \\ & \Delta \ell=2 m, \end{aligned}$

Change in volume $=A \Delta \ell=400 \mathrm{~cm}^3$

$M=400 \mathrm{~g} ;\left(\because \rho=1 \mathrm{~g} / \mathrm{cm}^3\right)$

$\begin{aligned} & \mathrm{f}_{\mathrm{c}}=\frac{\mathrm{v}}{4 \ell_1} \quad \mathrm{f}_{\mathrm{o}}=\frac{\mathrm{v}}{2 \ell_2} \\ & \left|\mathrm{f}_{\mathrm{c}}-\mathrm{f}_0\right|=7 \\ & \frac{\mathrm{v}}{4 \times 150}-\frac{\mathrm{v}}{2 \times 350}=7 \\ & \frac{\mathrm{v}}{600 \mathrm{~cm}}-\frac{\mathrm{v}}{700 \mathrm{~cm}}=7 \\ & \frac{\mathrm{v}}{6 \mathrm{~m}}-\frac{\mathrm{v}}{7 \mathrm{~m}}=7 \\ & \mathrm{v}\left(\frac{1}{42}\right)=7 \\ & \mathrm{v}=42 \times 7 \\ & =294 \mathrm{~m} / \mathrm{s} \end{aligned}$

To find the speed of the transverse waves produced in the string, we can use the formula for the fundamental frequency of a vibrating string:

$f = \frac{1}{2L} \cdot v$

where $f$ is the fundamental frequency, $L$ is the length of the string, and $v$ is the speed of the transverse waves.

First, we are given the mass of the string ($m = 18g$) and the linear mass density ($\mu = 20g/m$). We can find the length of the string by dividing the mass by the linear mass density:

$L = \frac{m}{\mu} = \frac{18g}{20g/m} = 0.9m$

Now we can plug in the values for the fundamental frequency ($f = 50Hz$) and the length of the string ($L = 0.9m$) into the formula:

$50Hz = \frac{1}{2(0.9m)} \cdot v$

To isolate $v$, we multiply both sides by $2(0.9m)$:

$v = 50Hz \cdot 2(0.9m) = 90 \mathrm{ms}^{-1}$

The speed of the transverse waves produced in the string is $90 ~\mathrm{ms}^{-1}$.

Alternate Method:

To find the speed of the transverse waves produced in the string, we can use the formula for the fundamental frequency of a vibrating string:

$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

where $f_1$ is the fundamental frequency, $L$ is the length of the string, $T$ is the tension in the string, and $\mu$ is the linear mass density of the string.

We're given that the fundamental frequency $f_1 = 50 ,\text{Hz}$, the mass of the string $m = 18 ,\text{g}$, and the linear mass density $\mu = 20 ,\text{g/m}$. To find the speed of the transverse waves, we need to find the tension $T$ and the length $L$ of the string.

First, let's find the length $L$ of the string using the mass and linear mass density:

$L = \frac{m}{\mu} = \frac{18 ,\text{g}}{20 ,\text{g/m}} = 0.9 ~\text{m}$

Now, we can rearrange the formula for the fundamental frequency to solve for the tension $T$:

$T = \mu \left(\frac{2Lf_1}{1}\right)^2$

Substitute the known values:

$T = 20 ,\text{g/m} \cdot \left(\frac{2 \cdot 0.9 ~\text{m} \cdot 50 ~\text{Hz}}{1}\right)^2$

$T = 20 ,\text{g/m} \cdot (90 ~\text{m/s})^2$

$T = 20 ,\text{g/m} \cdot 8100 ~\text{m}^2/\text{s}^2$

$T = 162000 ,\text{g m}/\text{s}^2$

Now, we can find the speed of the transverse waves $v$ using the formula:

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

Substitute the known values:

$v = \sqrt{\frac{162000}{20}}$

$v = \sqrt{8100} = 90~ \text{m/s}$

The speed of the transverse waves produced in the string is $90 ~\text{m/s}$.

We can use the fact that the ratio of frequencies is equal to the square root of the ratio of tensions:

$\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}$

In the first case, the mass attached to the string is $180 \mathrm{~g}$ and the frequency is $30 \mathrm{~Hz}$, so we have:

$\frac{f_2}{30~\mathrm{Hz}}=\sqrt{\frac{T_2}{T_1}}$

In the second case, the frequency is $50 \mathrm{~Hz}$, so we have:

$\frac{50~\mathrm{Hz}}{30~\mathrm{Hz}}=\sqrt{\frac{T_2}{T_1}}$

Simplifying, we get:

$\frac{5}{3}=\sqrt{\frac{T_2}{T_1}}$

Squaring both sides, we get:

$\frac{25}{9}=\frac{T_2}{T_1}$

Since the tension in the string is proportional to the mass attached to it, we can write:

$\frac{m}{180~\mathrm{g}}=\frac{T_2}{T_1}=\frac{25}{9}$

Solving for $m$, we get:

$m=\frac{25}{9}(180~\mathrm{g})=\boxed{500~\mathrm{g}}$

Therefore, the mass attached to the string in the second case is $500 \mathrm{~g}$.

Given that the first three resonance frequencies are in the ratio of $1:3:5$, we can express them as follows:

$f_1 = kf$ $f_3 = 3kf$ $f_5 = 5kf$

Where $k$ is a constant and $f_1, f_3$, and $f_5$ are the first, third, and fifth resonance frequencies, respectively. We are given that the frequency of the fifth harmonic is $405 \mathrm{~Hz}$, so we can write:

$f_5 = 5kf = 405 \mathrm{~Hz}$

Now we can solve for the constant $k$:

$k = \frac{405}{5} = 81 \mathrm{~Hz}$

We also know that the speed of sound in air is $v = 324 \mathrm{~ms}^{-1}$. The relationship between the speed of sound, the frequency, and the wavelength of a standing wave in a closed pipe can be expressed as follows:

$v = f\lambda$

Where $\lambda$ is the wavelength of the wave. For the first harmonic in a closed pipe, the length of the pipe is equal to one-fourth of the wavelength:

$L = \frac{1}{4}\lambda$

We can now substitute the expression for the wavelength in terms of the length into the equation for the speed of sound:

$v = f_1 \cdot 4L$

Now, we can substitute the value of $f_1 = kf = 81 \mathrm{~Hz}$ and the speed of sound $v = 324 \mathrm{~ms}^{-1}$ into the equation:

$324 = 81 \times 4L$

Now we can solve for the length of the organ pipe $L$:

$L = \frac{324}{81 \times 4} = \frac{324}{324} = 1 \mathrm{~m}$

The length of the organ pipe is $1 \mathrm{~m}$.

Given the wave equation:

$Y = 10^{-2} \sin 2 \pi(160t - 0.5x + \pi/4)$

Comparing this equation with the general form:

$Y = A \sin(2\pi(ft - kx + \phi))$

We can identify the wave number $k = 0.5\,\mathrm{m}^{-1}$ and the frequency $f = 160\,\mathrm{Hz}$. The wave speed $v$ can be found using the relationship between wave number, wave speed, and frequency:

$v = \frac{\omega}{k} = \frac{2\pi f}{2\pi k}$

Now, we can calculate the wave speed:

$v = \frac{2\pi \times 160}{2\pi \times 0.5} = \frac{160}{0.5}\,\mathrm{m/s}$

$v = 320\,\mathrm{m/s}$

Now, we need to convert the wave speed from meters per second to kilometers per hour:

$v = 320 \frac{\mathrm{m}}{\mathrm{s}} \times \frac{1\,\mathrm{km}}{1000\,\mathrm{m}} \times \frac{3600\,\mathrm{s}}{1\,\mathrm{h}}$

$v = 320 \times \frac{1}{1000} \times 3600\,\mathrm{km/h}$

$v = 1152\,\mathrm{km/h}$

So, the speed of the wave is $1152\,\mathrm{km/h}$.

The general equation for a transverse harmonic wave on a string is given by:

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

where $A$ is the amplitude of the wave, $k$ is the wave number, $\omega$ is the angular frequency, and $\phi$ is the phase constant. The wave velocity $v$ is related to the wave number and angular frequency by the formula:

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

Comparing the given equation with the general equation, we can see that:

$ A = 5 \, \text{cm} $

$ k = 0.003 \, \text{cm}^{-1} $

$ \omega = 6 \, \text{rad/s} $

Therefore, the wave velocity is:

$ v = \frac{\omega}{k} = \frac{6}{0.003} = 2000 \, \text{cm/s} = \boxed{20 \, \text{m/s}} $

The fundamental frequency (also known as the first harmonic) of a vibrating string is given by the formula:

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

where:

• (f) is the frequency,
• (v) is the speed of the wave in the string, and
• (L) is the length of the string.

In this case, the speed of the wave in the string stays the same because it depends on the properties of the string and the tension in it, which we can assume to be constant.

We can write the equation for the fundamental frequency of the original string and the shorter string:

$f_1 = \frac{v}{2L_1}$

$f_2 = \frac{v}{2L_2}$

where:

• $(f_1 = 120 \, \text{Hz})$ and $ (L_1 = 90 \, \text{cm})$ for the original string, and
• $(f_2 = 180 \, \text{Hz})$ and $(L_2)$ is what we're trying to find for the shorter string.

We can set up a ratio of these two equations:

$\frac{f_1}{f_2} = \frac{L_2}{L_1}$

Substituting in the given values, we get:

$\frac{120 \, \text{Hz}}{180 \, \text{Hz}} = \frac{L_2}{90 \, \text{cm}}$

Solving for ($L_2$) gives:

$L_2 = 90 \, \text{cm} \times \frac{120 \, \text{Hz}}{180 \, \text{Hz}} = 60 \, \text{cm}$

So, the length of the string producing a fundamental frequency of 180 Hz will be 60 cm.

An organ pipe that is open at both ends resonates at all harmonics, including the fundamental (first harmonic), second harmonic, third harmonic, etc.

The frequency $f$ of the $n$-th harmonic for a pipe open at both ends is given by:

$f_n = \frac{n v}{2L}$,

where:

• $n$ is the number of the harmonic,
• $v$ is the speed of sound, and
• $L$ is the length of the pipe.

To find the frequency of the second harmonic ($n = 2$), we can substitute the given values into the formula:

$f_2 = \frac{2 \times 360}{2 \times 0.4} = 900 \, \text{Hz}$.

Therefore, the frequency of the second harmonic is $900 \, \text{Hz}$.

$ \begin{aligned} & \text { Frequency of reflected sound }=\left(\frac{v+v_{\mathrm{c}}}{v-v_c}\right) f_0 \\\\ & f=\left(\frac{330+15}{330-15}\right) \times f_0 \\\\ & =\frac{345}{315} f_0 \\\\ & \frac{345}{315} f_0-f_0=40 \\\\ & \frac{30}{315} f_0=40 \\\\ & f_0=\frac{4 \times 315}{3}=420 \mathrm{~Hz} \end{aligned} $

Given, $y_{1}=10 \sin \left(\omega t+\frac{\pi}{3}\right) \mathrm{cm}$ and

$y_{2}=5(\sin \omega t+\sqrt{3} \cos \omega t)$

$ =10 \sin \left(\omega t+\frac{\pi}{3}\right) $

Thus the phase difference between the waves is 0 .

$ \text { so } A=A_{1}+A_{2}=20 \mathrm{~cm} $

$A_{R}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi}$

$ \begin{aligned} & 8=\sqrt{8^{2}+8^{2}+2 \times 8 \times 8 \cos \phi} \\\\ & \Rightarrow \cos \phi=-\frac{1}{2} \\\\ & \Rightarrow \phi=120^{\circ} \end{aligned} $

$f=f_{0}\left(\frac{v}{v-v_{s}}\right)$

$ \begin{aligned} & f=320\left(\frac{330}{330-66}\right) \\\\ & =320 \times \frac{330}{264} \\\\ & =400 \mathrm{~Hz} . \end{aligned} $

$ \begin{aligned} & \Delta x=\frac{\lambda}{2 \pi} \times\left(\frac{\pi}{3}\right)=\left(\frac{\lambda}{6}\right) \\\\ & \Rightarrow \quad \frac{\lambda}{6}=6 \mathrm{~m} \\\\ & \quad \lambda=36 \mathrm{~m} \\\\ & U=f\lambda=500 \mathrm{~Hz} \times 36 \\\\ & =18000 \mathrm{~m} / \mathrm{s} \\\\ & =18 \mathrm{~km} / \mathrm{s} \end{aligned} $

${v_s} = 36 \times {5 \over {18}} = 10$ m/sec

$f = {{v + {v_s}} \over {v - {v_s}}}{f_0}$

$ = {{340} \over {320}} \times 320$

$ = 340$ Hz

${v \over {2l}} = 50$ Hz

$ \Rightarrow T = {\left[ {100 \times \left( {{{30} \over {100}}} \right)} \right]^2} \times \mu $

$ \Rightarrow \mu = {{2700} \over {900}} = 3$

$100 = {v_0}{v \over {v - {v_c}}}$

$50 = {v_0}{v \over {v + {v_c}}}$

$2 = {{v + {v_c}} \over {v - {v_c}}}$

$2v - 2{v_c} = v + {v_c}$

${v_c} = {v \over 3}$

$100 = {v_0}{{v \times 3} \over {2v}} \Rightarrow {v_0} = {{200} \over 3} = {x \over 3}$

$ \Rightarrow x = 200$

${A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } $

$\sqrt 3 A = \sqrt {{A^2} + {A^2} + 2{A^2}\cos \phi } $

$3{A^2} = 2{A^2} + 2{A^2}\cos \phi $

$\cos \phi = {1 \over 2}$

$\phi = 60^\circ $

${f_1} = {f_0}\left( {{{320 - 5} \over {320}}} \right) = 640\left( {{{315} \over {320}}} \right)$

$ = 630$ Hz

${f_3} = {f_0}$ [No relative motion]

${f_2} = {f_0}\left[ {{{320 + 5} \over {320}}} \right] = 640\left( {{{325} \over {320}}} \right)$

$ = 650$

Beat frequency $ = {f_2} - {f_1}$

$ = 650 - 630 = 20$ Hz

Here, the apparent frequency is given by

$ \begin{aligned} &f^{\prime}=f\left(\frac{V-V_{0}}{V+V_{s}}\right)=720\left(\frac{(340+20)-20}{(340+20)-0}\right) \\\\ &=\frac{720 \times 340}{360}=680 \mathrm{~Hz} \end{aligned} $

$400 = {v \over {4({L_1} + e)}}$ ..... (i)

$400 = {{5v} \over {4({L_2} + e)}}$ ..... (ii)

$ \Rightarrow {L_1} + e = {\lambda \over 4} = 21$ cm

${L_2} + e = {{5\lambda } \over 4} = 105$ cm

$\Rightarrow$ e = 1 cm & L₂ = 104 cm

Given $340 = {n \over {4 \times 125}}v$

$ \Rightarrow n = 5$

So $\lambda = 100$ cm

So minimum height is ${\lambda \over 2} = 50$ cm

Given ${v_{20}} = 2{v_1}$

Also ${v_{20}} = 4 \times 19 + {v_1}$

So ${v_{20}} = 152\,Hz$

$2 \times \left( {{V \over {2{L_0}}}} \right) = \left( {{V \over {4{L_c}}}} \right)$

$ \Rightarrow {L_0} = 4{L_c}$

$ = 4 \times 20$

$ = 80$ cm

$A = \left| {10\cos (\pi x)} \right|$

At $x = {4 \over 3}$

$A = \left| {10\cos \left( {\pi \times {4 \over 3}} \right)} \right|$

$ = $ | $-$ 5 cm |

$\therefore$ Amp = 5 cm

$\mu = 9.0 \times {10^{ - 4}}{{kg} \over m}$

T = 900 N

$V = \sqrt {{T \over \mu }} = \sqrt {{{900} \over {9 \times {{10}^{ - 4}}}}} = 1000$ m/s

f₁ = 500 Hz

f = 550

${{nV} \over {2l}} = 500$ .... (i)

${{(n + 1)V} \over {2l}} = 500$ .... (ii)

(ii) (i) ${V \over {2l}} = 50$

$l = {{1000} \over {2 \times 50}} = 10$

${\lambda \over 4}$ = l $\Rightarrow$ $\lambda$ = 4l

f = ${V \over \lambda } = {V \over {4l}}$

$\Rightarrow$ 250 = ${{340} \over {4l}}$

$\Rightarrow$ l = ${{34} \over {4 \times 25}}$ = 0.34 m

l = 34 cm

Image

V_x = 36 km/hr = 10 m/s

V_y = 72 km/hr = 20 m/s

by doppler's effect

$F' = {F_0}\left( {{{V \pm {V_0}} \over {V \pm {V_s}}}} \right)$

$1320 = {F_0}\left( {{{340 + 20} \over {340 - 10}}} \right) \Rightarrow {F_0} = 1210$ Hz

y₁ = A₁ sin k(x $-$ vt)

y₁ = 12 sin 6.28 (x $-$ vt)

y₂ = 5 sin 6.28 (x $-$ vt + 3.5)

$\Delta \phi = {{2\pi } \over \lambda }(\Delta x)$

$ = K(\Delta x)$

$ = 6.28 \times 3.5 = {7 \over 2} \times 2\pi = 7\pi $

${A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } $

${A_{net}} = \sqrt {{{(12)}^2} + {{(5)}^2} + 2(12)(5)\cos (7\pi )} $

$ = \sqrt {144 + 25 - 120} $

Image

$f' = f\left( {{{C - {V_0}} \over {C + {V_s}}}} \right)$

$ \Rightarrow $ $1800 = f\left( {{{340 - 20} \over {340 + 20}}} \right)$

f = 2025 Hz

For node

cos(1.57 cm^$-$1)x = 0

(1.57 cm^$-$1)x = ${\pi \over 2}$

x = ${\pi \over {2(1.57)}}$ cm = 1 cm

$\because$ Since, frequency received by the wall,

$f' = \left( {{{{v_s}} \over {{v_s} - v}}} \right){f_o}$ .... (i)

where, v_s = velocity of sound in air, v = velocity of car and f_o = observed frequency of sound.

Reflected frequency received by man is

$f'' = \left( {{{{v_s} + v} \over {{v_s}}}} \right)f'$ ..... (ii)

From Eqs. (i) and (ii), we get

$f'' = \left( {{{{v_s} + v} \over {{v_s}}}} \right)\left( {{{{v_s}} \over {{v_s} - v}}} \right){f_o} \Rightarrow f'' = \left( {{{{v_s} + v} \over {{v_s} - v}}} \right){f_o}$

$ \Rightarrow 500 = \left( {{{330 + v} \over {330 - v}}} \right) \times 400 \Rightarrow {{500} \over {400}} = {{(330 + v)} \over {(330 - v)}}$

$ \Rightarrow 5(330 - v) = 4(330 + v) \Rightarrow 1650 - 5v = 1320 + 4v$

$ \Rightarrow 9v = 330 \Rightarrow v = {{330} \over 9}$ m/s

or $v = {{330} \over 9} \times {{18} \over 5} = 132$ km/h

As per question,

at t = 0, y = ${1 \over {{{(1 + x)}^2}}}$

and at t = 1 s, y = ${1 \over {1 + {{(x - 2)}^2}}}$ .... (i)

As we know,

At t = t s, y = ${1 \over {1 + {{(x - vt)}^2}}}$

So, at t = 1 s, y = ${1 \over {1 + {{(x - v)}^2}}}$ .... (ii)

On comparing Eqs. (i) and (ii), we get

v = 2 ms^$-$1

Hence, the velocity of the wave will be 2 m/s.

First overtone of open pipe $ = {{{v_2}} \over {{L_2}}}$

First overtone of closed pipe at one end $ = {{3v} \over {4L}}$

As per question,

${{3V} \over {4L}} = {{{V_2}} \over L}$

$ \Rightarrow \sqrt {{B \over {{\rho _1}}}} \,.\,{3 \over {4L}} = \sqrt {{B \over {{\rho _2}}}} \,.\,{1 \over {{L_2}}}$ ($\because$ $V = \sqrt {{B \over \rho }} $)

$ \Rightarrow {L_2} = {{4L} \over 3}\sqrt {{{{\rho _1}} \over {{\rho _2}}}} $ ..... (i)

According to question, the length of the open pipe is

${x \over 3}L\sqrt {{{{\rho _1}} \over {{\rho _2}}}} $ ..... (ii)

Comparing Eqs. (i) and (ii), we get

$x = 4$

$\mu = 0.135$ gm/cm

$\mu = 0.135 \times {{{{10}^{ - 3}}} \over {{{10}^{ - 2}}}}{{kg} \over m}$

y = $-$0.21 sin (x + 30t)

$v = {\omega \over K} = {{30} \over 1}$ = 30 m/s

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

T = v² $\times$ $\mu$

T = (30)² $\times$ 0.135 $\times$ 10^$-$1

T = 900 $\times$ 0.135 $\times$ 10^$-$1

T = 12.15 N

T = 1215 $\times$ 10^$-$2 N

x = 1215

Speed of transverse wave is

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

$\ln V = {1 \over 2}\ln T - {1 \over 2}\ln \mu $

${{\Delta V} \over V} = {1 \over 2}{{\Delta T} \over T}$

$ = {1 \over 2} \times 4$

$ \Rightarrow $ ${{\Delta V} \over V} = 2\% $

Given, v_A = v_B = 7.2 kmh^$-$1

$ = {{72} \over {10}} \times {5 \over {18}}$ = 2 ms^$-$1

Frequency of source, f_s = 676 Hz

Speed of sound in air, v = 340 ms^$-$1

Let f₀ be the frequency heard by each driver.

By using Doppler effect for A,

$(v - {v_A}){f_s} = (v + {v_B}){f_0}$

$ \Rightarrow {f_0} = \left( {{{v + {v_A}} \over {v - {v_B}}}} \right){f_s} = \left( {{{340 + 2} \over {340 - 2}}} \right)676 = {{342} \over {338}} \times 676 = 684$ Hz

Now, beat frequency $ = {f_0} - {f_s} = 684 - 676 = 8$ Hz

Velocity of sound v = $\sqrt {{B \over \rho }} $

$ \therefore $ ${{{v_{pipe}}} \over {{v_{air}}}} = \sqrt {{{{B \over {2\rho }}} \over {{B \over \rho }}}} $

$ \Rightarrow $ v_pipe = ${{{{v_{air}}} \over {\sqrt 2 }}}$ = ${{{300} \over {\sqrt 2 }}}$ = 150${\sqrt 2 }$

We know, for open organ pipe

f_n = $\left( {n + 1} \right){{{v_{pipe}}} \over {2l}}$

Difference between 2^nd and fundamental frequency

f₁ - f₀ = ${{{v_{pipe}}} \over {2l}}$ = ${{150\sqrt 2 } \over {2\left( 1 \right)}}$ = 106.06 Hz