Waves

Velocity of point $p=dy/dt$

= $-$(velocity of wave $\times$ $dy/dx$)

Here, $dy/dx=$ negative ($-$ve). Therefore, velocity at point P is positive and is along X-axis only.

Equation of a wave moving in positive X-axis is given as,

$y = A\sin (wt - \phi )$ .... (i)

or, ${V_P} = Aw\cos (wt - \phi )$ .... (ii)

Here, $y' = 5$ cm, $A = 10$ cm

$\therefore$ $5 = 10\sin (wt - \phi ) \Rightarrow wt - \phi = 30^\circ $ .... (iii)

Put equation (iii) in equation (ii),

${V_P} = 0.10 \times w\cos 30^\circ $

Now, $V = {V_\lambda }$, $\therefore$ $v = V/\lambda = {{0.10} \over {0.5}} = 0.2$

$\therefore$ $w = 2\pi v = 2\pi \times 0.2 = 0.4\pi $

$ \Rightarrow {V_P} = 0.1 \times 0.4\pi \times {{\sqrt 3 } \over 2} = {{\sqrt 3 } \over {50}}\pi \,\widehat j$ m/s

It is in positive Y-direction.

Let f be frequency of string

According to close open pipe, we have

$ \Rightarrow f = {{3V} \over {4l}} = {{3 \times 340} \over {4 \times 75 \times {{10}^{ - 2}}}}$

$ = 340$ Hz

Where, $V = 340$ m/s

$l = 75$ cm $ \times {10^{ - 2}}$ m

$ = 75 \times {10^{ - 2}}$ m

And, n = frequency of tuning fork

$ = (340 + 4)$ or $(340 - 4)$

$ \Rightarrow n = 344$ or 336

As tension T increase, beat frequency decrease

But frequency of tuning fork increases

$\therefore$ $n = 344$ Hz

Wave produced by tuning Fork in a vertical plane as shown in above figure.

Therefore, fringes of tuning Fork is kept in vertical plane.

$v_{\mathrm{SA}}=(340+20) \mathrm{m} / \mathrm{s}=360 \mathrm{~m} / \mathrm{s} $

$v_{\mathrm{SB}}=(340-30) \mathrm{m} / \mathrm{s}=310 \mathrm{~m} / \mathrm{s} $

where, speed of sound in air = 340 m/s

Hence, the speed of sound depends on the frame of reference of observer.

Since all the passengers in train A are moving with a velocity of $20 \mathrm{~m} / \mathrm{s}$ therefore the distribution of sound intensity of the whistle by the passengers in the train $A$ is uniform.

Trick : Total frequency $=(1120-800) \mathrm{Hz} =320 \mathrm{~Hz}$

There is no relative motion between source and observer.

Therefore, all the passengers will hear the sound of same intensity.

Frequency $(f') = {f_1}\left( {{{v - {v_0}} \over {v - {v_s}}}} \right) = 800\left[ {{{340 - 30} \over {340 - 20}}} \right]$

$=800 \times \frac{31}{32}$ ..... (i)

where, $\left(\frac{v-v_{0}}{v-v_{\mathrm{S}}}\right)=\frac{31}{32}$ ..... (ii)

Frequency $\left(f^{\prime}\right)=f_{2}\left(\frac{v-v_{0}}{v-v_{\mathrm{S}}}\right)=1120 \times \frac{31}{32}$ ..... (iii)

$\therefore \quad f^{\prime}-f^{\prime}=(1120-800) \frac{31}{32}=320 \times \frac{31}{32}$

$=310 \mathrm{~Hz}$

Reason:

In above figure shown that S.H.M

Therefore, $f_s=Kx$

Given, $v = {c_1}\sqrt {{c_2} - {x^2}} $ ..... (i)

For a simple harmonic motion,

$v = \omega \sqrt {{A^2} - {x^2}} $ ..... (ii)

On comparing eq. (i) & (ii) therefore it shows the S.H.M of particle.

A $\to$ P

(b) Given,

$v=-kx$

At $x=0,v=0$

$x=2,v=-2k$

$x=-5,v=+5k$

The above consideration show that particle is coming from minus x-axis.

Velocity is decreasing with time.

Graph:

Therefore, no change in direction, velocity keeps on decreasing. So that kinetic energy is decreasing with time.

B $\to q_1r$

(C) From the reference frame of the observer in the elevator a constant force acts on the object. Hence the motion of the object will remain simple harmonic.

Therefore, option 1 is correct.

Velocity of object

$v = 2\sqrt {G{{{M_e}} \over {{R_e}}}} = \sqrt 2 \sqrt {2G{{{M_e}} \over {{R_e}}}} = \sqrt 2 {v_e}$

Since, $v > {v_e}$ (where $v_e$ is escape velocity) object will move out of earth.

Hence, option (Q) and (R) are correct.

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} $

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: $y = 0.1 \sin(30t + 1.5x \pm \phi)$