Waves

Frequency of $1^{\text {st }}$ harmonic of $\mathrm{AB}=\frac{1}{2 l} \sqrt{\frac{\mathrm{~T}_{\mathrm{AB}}}{m}}$

Frequency of $2^{\text {nd }}$ harmonic of $\mathrm{CD}=\frac{1}{l} \sqrt{\frac{\mathrm{~T}_{\mathrm{CD}}}{m}}$

Given that 2 frequencies are equal.

$ \begin{aligned} \frac{1}{2 l} \sqrt{\frac{\mathrm{~T}_{\mathrm{AB}}}{m}} & =\frac{1}{l} \sqrt{\frac{\mathrm{~T}_{\mathrm{CD}}}{m}} \\ \frac{\mathrm{~T}_{\mathrm{AB}}}{m} \times \frac{1}{4 l^2} & =\frac{1}{l^2} \cdot \frac{\mathrm{~T}_{\mathrm{CD}}}{m} \quad(\text { On squaring }) \\ \frac{\mathrm{T}_{\mathrm{AB}}}{4} & =\mathrm{T}_{\mathrm{CD}} \\ \mathrm{~T}_{\mathrm{AB}} & =4 \mathrm{~T}_{\mathrm{CD}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

For rotational equilibrium of massless rod, Taking torque about point D ,

$ \mathrm{T}_{\mathrm{AB}} \times x=\mathrm{T}_{\mathrm{CD}}(\mathrm{~L}-x) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

For translational equillibrium,

$ \mathrm{T}_{\mathrm{AB}}+\mathrm{T}_{\mathrm{CD}}=m g \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

From (i) and (iii), $4 \mathrm{~T}_{\mathrm{CD}}+\mathrm{T}_{\mathrm{CO}}=m g$

$ \begin{array}{ll} \Rightarrow & \mathrm{T}_{\mathrm{CD}}=\frac{m g}{5} \\ \therefore & \mathrm{~T}_{\mathrm{AB}}=4 \times \frac{m g}{5}=\frac{4 m g}{5} \end{array} $

On substituting above value in (i) we get

$ \Rightarrow \quad \begin{aligned} \frac{4 m g}{5} \times x & =\frac{m g}{5}(L-x) \\ 4 x & =\mathrm{L}-x \\ 5 x & =\mathrm{L} \\ x & =\mathrm{L} / 5 . \end{aligned} $

Number of times the intensity is maximum on superposition in 1 , is calculated as follows:

Here,

$ \begin{aligned} y_1 & =\mathrm{A} \cos (0.5 \pi x-100 \pi t) \\ y_1 & =\mathrm{A} \cos \left(\frac{2 \pi}{\lambda} x-2 \pi v_1 t\right) \\ y_2 & =\mathrm{A} \cos (0.46 \pi-92 \pi t) \\ & =\mathrm{A} \cos \frac{2 \pi}{\lambda^2} x-2 \pi v_2 t \end{aligned} $

The two waves travelling in same direction will form beats having frequency $=$ frequency of maximum

$ =\mathrm{v}_1=\mathrm{v}_2=50-46=4 \mathrm{~Hz} \text { or } 4 \mathrm{~s}^{-1} . $

$\left[\right.$ Here, $2 \pi v_1 t=100 \pi t$ and $2 \pi v_2 t=92 \pi, v_2=46 \mathrm{~Hz}$ ]

Comparing the louder wave with simple harmonic waves,

waves velocity is $\frac{w}{k}=\frac{100}{0.5}=200 \mathrm{~m} / \mathrm{s}$.

Average frequency of wave after superposition will be

$ f=\frac{f_1+f_2}{2}=\frac{100+92}{2}=96 \mathrm{~Hz} $

$n' = n\left[ {{{v + {v_0}} \over v}} \right] = n\left[ {{{v + {v \over 5}} \over v}} \right] = n\left[ {{6 \over 5}} \right]$

${{n'} \over n} = {6 \over 5};{{n' - n} \over n}$

$ = {{6 - 5} \over 5} \times 100 = 20\% $

No. of beats heard when fork $2$ is sounded with fork $1$ $ = \Delta n = 4$

Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from $4$ to $6$ in this case) then the frequency of the unknown fork $2$ is given by,

$n = {n_0} - \Delta n = 200 - 4 = 196Hz$

Let the speed of the train be V$_t$,

Let $v$ be the actual frequency of the whistle.

Speed of sound $\mathrm{V}_{s}=300 \mathrm{~m} / \mathrm{s}$

While train is approaching, $v_{2}=2.2 \mathrm{~kHz}$

$=2.2\times10^3$ Hz

$v_{2}=\frac{\mathrm{V}_{s}}{\mathrm{~V}_{s}-\mathrm{V}_{t}} \times v$

$2.2 \times 10^{3}=\frac{\mathrm{V}_{s}}{\mathrm{~V}_{s}-\mathrm{V}_{t}} \times v=\frac{300}{300-\mathrm{V}_{t}} \times v$ ...... (i)

While train is receding,

$v_1=1.8$ kHz = $1.8\times10^3$ Hz

${v_1} = {{{V_s}} \over {{V_s} + {V_t}}} \times v$

$1.8 \times {10^3} = {{300} \over {300 + {V_t}}} \times v$ .... (ii)

Dividing equation (i) by (ii)

${{2.2 \times {{10}^3}} \over {1.8 \times {{10}^3}}} = {{{{300} \over {300 - {V_t}}} \times v} \over {{{300} \over {300 + {V_t}}} \times v}}$

$ = {{2.2} \over {1.8}} = {{300 + {V_t}} \over {300 - {V_t}}}$

$660 - 2.2\,{V_t} = 540 + 1.8\,{V_t}$

$2.2\,{V_t} + 1.8\,{V_t} = 660 - 540$

$4{V_t} = 120$

${V_t} = 30$ m/s

Step 1: Understanding Symbols

Let $\omega$ be the angular velocity and $A$ be the amplitude.

Step 2: Maximum Velocity

The highest speed up and down (maximum transverse velocity) is $V_{\max} = \omega A = 3~\mathrm{m/s}$.

Step 3: Maximum Acceleration

The highest acceleration up and down is $\omega^2 A = 90~\mathrm{m/s}^2$.

Step 4: Wave Velocity Relation

The speed that the wave itself travels is $\frac{\omega}{k} = 20~\mathrm{m/s}$, where $k$ is called the wave number.

Step 5: Find Angular Velocity $\omega$

To find $\omega$, divide the maximum acceleration equation by the maximum velocity equation:

$\frac{A \omega^2}{A \omega} = \frac{90}{3}$

So, $\omega = 30~\mathrm{rad/s}$.

Step 6: Find Amplitude $A$

From the maximum velocity equation: $A = \frac{3}{30} = 0.1~\mathrm{m}$.

Step 7: Find Wave Number $k$

We know $k = \frac{\omega}{v}$, where $v$ is the wave speed.

$k = \frac{30}{20} = 1.5~\mathrm{m}^{-1}$

Step 8: Write the Wave Equation

The wave can be written as: $y = A \sin(\omega t + k x + \phi)$, where $\phi$ is a starting phase (it can take any value).

So, the waveform is: $y = 0.1 \sin(30t + 1.5x \pm \phi)$

From equation given,

$\omega = 100$ and $k = 20,$ $v = {\omega \over k} = {{100} \over {20}} = 5m/s$

A tuning fork of frequency $256$ $Hz$ makes $5$ beats/ second with the vibrating string of a piano. Therefore the frequency of the vibrating string of piano is $\left( {256 \pm 5} \right)$ $Hz$ ie either $261$$Hz$ or $251$ $Hz.$ When the tension in the piano string increases, its frequency will increases. Now since the beat frequency decreases, we can conclude that the frequency of piano string is $251$ $Hz$

KEY CONCEPT : For a string vibrating between two rigid support, the fundamental frequency is given by

$n = {1 \over {2\ell }}\sqrt {{T \over \mu }} = {1 \over {2 \times }}\sqrt {{{10 \times 9.8} \over {9.8 \times {{10}^{ - 3}}}}} = 50Hz$

As the string is vibrating in resonance to a.c of frequency $n,$ therefore both the frequencies are same.

$y = {10^{ - 4}}\sin \left( {600t - 2x + {\pi \over 3}} \right)$

But $y = A\sin \left( {\omega t - kx + \phi } \right)$

On comparing we get $\omega = 600;\,k = 2$

$v = {\omega \over k} = {{600} \over 2} = 300\,m{s^{ - 1}}$

This will happen for fundamental mode of vibration as shown in the figure. ${S_1}$ and ${S_2}$ are rigid support

Here ${\lambda \over 2} = 40\,\,\,\,\,\,$ $\therefore$ $\lambda = 80\,cm$

To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of $\pi $. The equation of the reflected wave will be

$y = a\sin \left( {\omega t + kx + \pi } \right)$

$ \Rightarrow y = - a\sin \left( {\omega t + kx} \right)$

KEY CONCEPT : The frequency of a tuning fork is given by the expression

$f = {{{m^2}k} \over {4\sqrt 3 \pi {\ell ^2}}}\sqrt {{Y \over \rho }} $

As temperature increases, $\ell $ increases and therefore $f$ decreases.

KEY CONCEPT : The fundamental frequency for closed organ pipe is given by ${\upsilon _c} = {v \over {4\ell }}$ and

For open organ pipe is given by ${\upsilon _0} = {v \over {2\ell }}$

$\therefore$ ${{{\upsilon _0}} \over {{\upsilon _c}}} = {v \over {2\ell }} \times {{4\ell } \over v} = {2 \over 1}$

A tuning fork produces $4$ beats/sec with another tuning fork of frequency $288$ cps. From this information we can conclude that the frequency of unknown fork is $288+4$ $cps$ or $288-4$ $cps$ i.e. $292$ $cps$ or $284$ $cps.$

Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork. Here also beats per second decreases to 2 from 4. So the difference between frequency decreases.

This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork.

So the frequency of the unknown tuninh fork is = 292 Hz