Waves — Page 6 | JEE Physics

Q51

A longitudinal wave is represented by

x = 10\sin 2\pi \left( {nt - {x \over \lambda }} \right)

cm. The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to :

A 2

\pi

B 5

\pi

C

\pi

D

{{5\pi } \over 2}

Correct Answer

Option B

Solution

Particle velocity =

{{\partial x} \over {\partial t}}

$\Rightarrow$ Maximum particle velocity

= (2\pi n)\,(10)

\Rightarrow (2\pi n)\,(10) = (n\lambda )\,(4)

\Rightarrow \lambda = 5\pi

Q52

The equations of two waves are given by : y1 = 5 sin 2

\pi

(x - vt) cm y2 = 3 sin 2

\pi

(x

-

vt + 1.5) cm These waves are simultaneously passing through a string. The amplitude of the resulting wave is :

A 2 cm

B 4 cm

C 5.8 cm

D 8 cm

Correct Answer

Option A

Solution

{y_1} = 5\sin (2\pi x - 2\pi vt)

{y_2} = 3\sin (2\pi x - 2\pi vt + 3\pi )

$\Rightarrow$ Phase difference = 3 $\pi$

\Rightarrow {A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos (3\pi )}

\Rightarrow {A_{net}} = 2

cm

Q53

Which of the following equations correctly represents a travelling wave having wavelength

\lambda

= 4.0 cm, frequency v = 100 Hz and travelling in positive x-axis direction?

A

y = A\sin [(0.50\,\pi \,c{m^{ - 1}})x - (100\,\pi \,{s^{ - 1}})t]

B

y = A\sin \,\,2\pi [(0.25\,\,c{m^{ - 1}})x - (50\,{s^{ - 1}})t]

C

y = A\sin \left[ {\left( {{{2\pi } \over 4}\,c{m^{ - 1}}} \right)x - \left( {{{2\pi } \over {100}}\,{s^{ - 1}}} \right)t} \right]

D

y = A\sin \,\pi [(0.5\,\,c{m^{ - 1}})x - (200\,\,{s^{ - 1}})t]

Correct Answer

Option D

Solution

We know, equation of wave travelling in positive x-direction is -

y = A\sin (kx - wt)

where

k = {{2\pi } \over \lambda }

and

w = 2\pi f

Here given $\lambda$ = 4 cm and frequency (f) = 100 Hz $\therefore$

k = {{2\pi } \over 4} = 0.5\pi

cm $-$ 1 and

w = 2\pi \times 100 = 200\pi

s $-$ 1 $\therefore$ Equation of travelling wave,

y = A\sin (0.5\pi x - 200\pi t)

Q54

A transverse wave is represented by

y=2 \sin (\omega t-k x)\, \mathrm{cm}

. The value of wavelength (in

\mathrm{cm}

) for which the wave velocity becomes equal to the maximum particle velocity, will be :

A 4

\pi

B 2

\pi

C

\pi

D 2

Correct Answer

Option A

Solution

{\omega \over k} = A\omega

\Rightarrow k = {1 \over A} = {1 \over {2\,cm}}

\Rightarrow {{2\pi } \over \lambda } = {1 \over {2\,cm}}

\Rightarrow \lambda = 4\pi \,cm

Q55

In the wave equation

y=0.5 \sin \dfrac{2 \pi}{\lambda}(400 \mathrm{t}-x) \,\mathrm{m}

the velocity of the wave will be:

A 200 m/s

B 200

\sqrt2

m/s

C 400 m/s

D 400

\sqrt2

m/s

Correct Answer

Option C

Solution

{v_{wave}} = \left| {{{coefficient\,of\,t} \over {coefficient\,of\,x}}} \right|

= {{400} \over 1} = 400

m/s

Q56

A steel wire with mass per unit length

7.0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}

is under tension of

70 \mathrm{~N}

. The speed of transverse waves in the wire will be:

A

10 \mathrm{~m} / \mathrm{s}

B

50 \mathrm{~m} / \mathrm{s}

C

100 \mathrm{~m} / \mathrm{s}

D

200 \pi\mathrm{~m} / \mathrm{s}

Correct Answer

Option C

Solution

Speed of transverse wave $=\sqrt{\dfrac{T}{\mu}}$ $=\sqrt{\dfrac{70}{7 \times 10^{-3}}}=100 \mathrm{~m} / \mathrm{s}$

Q57

A person observes two moving trains, 'A' reaching the station and 'B' leaving the station with equal speed of

30 \mathrm{~m} / \mathrm{s}

. If both trains emit sounds with frequency

300 \mathrm{~Hz}

, (Speed of sound:

330 \mathrm{~m} / \mathrm{s}

) approximate difference of frequencies heard by the person will be:

A 10 Hz

B 55 Hz

C 80 Hz

D 33 Hz

Correct Answer

Option B

Solution

By doppler effect : $f^{\prime}=f_{0}\left[\dfrac{v-v_{0}}{v-v_{s}}\right]$ $\Rightarrow \quad f_{A}^{\prime}=300\left[\dfrac{330}{330-30}\right] \mathrm{Hz}$

=330 \mathrm{~Hz}

And $f_{B}^{\prime}=300\left[\dfrac{330}{330+30}\right] \mathrm{Hz}$

=\frac{5}{6} \times 330 \mathrm{~Hz}=275 \mathrm{~Hz}

$\therefore \Delta \mathrm{f}=330-275=55 \mathrm{~Hz}$ .

Q58

A car P travelling at

20 \mathrm{~ms}^{-1}

sounds its horn at a frequency of

400 \mathrm{~Hz}

. Another car

\mathrm{Q}

is travelling behind the first car in the same direction with a velocity

40 \mathrm{~ms}^{-1}

. The frequency heard by the passenger of the car

\mathrm{Q}

is approximately [Take, velocity of sound

=360 \mathrm{~ms}^{-1}

]

A 485 Hz

B 514 Hz

C 421 Hz

D 471 Hz

Correct Answer

Option C

Solution

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.

Q59

For a periodic motion represented by the equation

y=\sin \omega \mathrm{t}+\cos \omega \mathrm{t}

the amplitude of the motion is

A 1

B

\sqrt2

C 0.5

D 2

Correct Answer

Option B

Solution

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 $t = \dfrac{\pi}{4\omega} + \dfrac{n\pi}{\omega}$ , 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 $\sqrt{2}$

Q60

The engine of a train moving with speed

10 \mathrm{~ms}^{-1}

towards a platform sounds a whistle at frequency

400 \mathrm{~Hz}

. The frequency heard by a passenger inside the train is: (neglect air speed. Speed of sound in air

=330 \mathrm{~ms}^{-1}

)

A 200 Hz

B 412 Hz

C 400 Hz

D 388 Hz

Correct Answer

Option C

Solution

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.