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Waves

NEET Physics · 100 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
A wave in a string has an amplitude of 2 cm. The wave travels in the +ve direction of x axis with a speed of 128 m/s. and it is noted that 5 complete waves fit in 4m length of the string. The equation describing the wave is
A y = (0.02) m sin (15.7 x −- 2010t)
B y = (0.02) m sin (15.7 x + 2010t)
C y = (0.02) m sin (7.85 x −- 1005t)
D y = (0.02) m sin (7.85 x + 1005t)
Correct Answer
Option C
Solution

As

A=2cm,ωk=128ms−1A = 2cm,{\omega \over k} = 128m{s^{ - 1}}

,

5λ=45\lambda = 4

,

λ=45m\lambda = {4 \over 5}m

y = A sin (kx – ω\omegat)

k=2πλ=2π×54=31.44=7.85k = {{2\pi } \over \lambda } = {{2\pi \times 5} \over 4} = {{31.4} \over 4} = 7.85

∴\therefore ω\omega = 128 × 7.85 = 1005 so y = 0.02 m sin (7.85x – 1005 t)

Q42
The wave described by y = 0.25 sin(10π\pi x −- 2π\pi t), where x and y are in metres and t in seconds, is a wave travelling along the
A +ve x direction with frequency 1 Hz and wavelength λ\lambda = 0.2 m.
B −-ve x direction with amplitude 0.25 m and wavelength λ\lambda = 0.2 m.
C −-ve x direction with frequency 1 Hz.
D +ve x direction with frequency π\pi Hz and wavelength λ\lambda = 0.2 m.
Correct Answer
Option A
Solution

y = 0.25sin(10π\pix – 2π\pit)

ymax⁡=0.25{y_{\max }} = 0.25
k=2πλ=10π⇒λ=0.2 mk = {{2\pi } \over \lambda } = 10\pi \Rightarrow \lambda = 0.2\,m
ω=2πf=2π⇒f=1 Hz\omega = 2\pi f = 2\pi \Rightarrow f = 1\,Hz

The sign is negative inside the bracket. Therefore this wave travels in the positive x-direction.

Q43
Two periodic waves of intensities II 1 and II 2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
A (I1−I2)2{\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}
B 2(I1+I2)2\left( {{I_1} + {I_2}} \right)
C I1+I2{{I_1} + {I_2}}
D (I1+I2)2{\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}
Correct Answer
Option B
Solution

Other factors such as ω\omega and v remaining the same, I = A 2 constant K, or A =

IK\sqrt {{I \over K}}

On superposition

Amax⁡=A1+A2{A_{\max }} = {A_1} + {A_2}

and

Amin⁡=A1−A2{A_{\min }} = {A_1} - {A_2}
∴Amax⁡2=A12+A22+2A1A2\therefore A_{\max }^2 = A_1^2 + A_2^2 + 2{A_1}{A_2}
⇒Imax⁡K=I1K+I2K+2I1I2K\Rightarrow {{{I_{\max }}} \over K} = {{{I_1}} \over K} + {{{I_2}} \over K} + {{2\sqrt {{I_1}{I_2}} } \over K}
Amin⁡2=A12+A22−2A1A2A_{\min }^2 = A_1^2 + A_2^2 - 2{A_1}{A_2}
⇒Imin⁡K=I1K+I2K−2I1I2K\Rightarrow {{{I_{\min }}} \over K} = {{{I_1}} \over K} + {{{I_2}} \over K} - {{2\sqrt {{I_1}{I_2}} } \over K}
∴Imax⁡+Imin⁡=2I1+2I2\therefore {I_{\max }} + {I_{\min }} = 2{I_1} + 2{I_2}
Q44
A point performs simple harmonic oscillation of period T and the equation of motion is given by x = a sin(ω\omega t + π\pi /6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
A T/3
B T/12
C T/8
D T/6
Correct Answer
Option B
Solution

We have

x=asin⁡(ωt+π6)x = a\sin \left( {\omega t + {\pi \over 6}} \right)

∴\therefore Velocity,

v=dxdt=aωcos⁡(ωt+π6)v = {{dx} \over {dt}} = a\omega \cos \left( {\omega t + {\pi \over 6}} \right)

Maximum velocity = aω\omega According to question,

aω2=aωcos⁡(ωt+π6){{a\omega } \over 2} = a\omega \cos \left( {\omega t + {\pi \over 6}} \right)
⇒cos⁡(ωt+π6)=12=cos⁡60o\Rightarrow \cos \left( {\omega t + {\pi \over 6}} \right) = {1 \over 2} = \cos {60^o}

or

cos⁡π3{\cos {\pi \over 3}}
⇒ωt=π3−π6⇒ωt=π6\Rightarrow \omega t = {\pi \over 3} - {\pi \over 6} \Rightarrow \omega t = {\pi \over 6}
⇒2πT.t=π6⇒t=T12\Rightarrow {{2\pi } \over T}.t = {\pi \over 6} \Rightarrow t = {T \over {12}}
Q45
Two vibrating tuning forks produce waves given by y 1 = 4 sin 500π\pi t and y 2 = 2 sin506 π\pi t. Number of beats produced per minute is
A 360
B 180
C 60
D 3
Correct Answer
Option B
Solution

Y 1 = 4 sin500π\pit, Y 2 = 2 sin506π\pit

ω1=500π{\omega _1} = 500\pi

,

ω1=2πυ{\omega _1} = 2\pi \upsilon

,

υ1=250,υ2=253{\upsilon _1} = 250,{\upsilon _2} = 253
υ=υ2−υ1=253−250=3\upsilon = {\upsilon _2} - {\upsilon _1} = 253 - 250 = 3

beats/s ∴\therefore Number of beats per minute = 3 × 60 = 180

Q46
The time of reverberation of a room A is one second. What will be the time (in seconds of reverberation of a room, having all the dimensions double of those of room A ?
A 1
B 2
C 4
D 1/2
Correct Answer
Option B
Solution

Reverberation time is defined as the time during which the intensity of sound in the auditorium becomes one millionth of the initial intensity.

Sabine has shown that standard reverberation time for an auditorium is given by the formula

TR=KVαS{T_R} = K{V \over {\alpha S}}

Here, V is volume of the auditorium, S is the surface area.

∴TR=K.VαS=1\therefore {T_R} = {{K.V} \over {\alpha S}} = 1

(given)

⇒TR=K.l36αl2=K6αl\Rightarrow {T_R} = {{K.{l^3}} \over {6\alpha {l^2}}} = {K \over {6\alpha }}l
TR=K.l36αl2=K6αl{T_R} = {{K.{l^3}} \over {6\alpha {l^2}}} = {K \over {6\alpha }}l

(Assuming auditorium to be cubic in shape)

∴TR∝l\therefore {T_R} \propto l

If dimension is doubled, reverberation time t will be doubled. So, New T R = 2 sec.

Q47
Two sound waves with wavelengths 5.0 m and 5.5. m respectively, each propagate in a gas with velocity 330 m/s. We expect the following number of beats per second.
A 6
B 12
C 0
D 1
Correct Answer
Option A
Solution

Frequencies of sound waves are

3305{{330} \over 5}

&

3305.5{{330} \over {5.5}}

i.e., 66 Hz and 60 Hz Frequencies of beat = 66 – 60 = 6 per second

Q48
Which one of the following statements is true ?
A both light and sound waves can travel in vaccum
B both light and sound waves in air are transverse
C The second waves in air are longitudinal while the light waves are transverse
D both light and sound waves in air are
Correct Answer
Option C
Solution

Light waves are electromagnetic waves.

Light waves are transverse in nature and do not require a medium to travel, hence they can travel in vacuum.

Sound waves are longitudinal waves and require a medium to travel.

They do not travel in vacuum.

Q49
A transverse wave propagating along x-axis is represented by y(x, t) = 8.0 sin (0.5 π\pi x −- 4π\pi t −- π\pi /4) where x is in metres and t is in seconds. The speed of the wave is
A 8 m/s
B 4π\pi m/s
C 0.5π\pi m/s
D π\pi /4 m/s.
Correct Answer
Option A
Solution
y(x,t)=8.0sin⁡(0.5πx−4πt−π4)y\left( {x,t} \right) = 8.0\sin \left( {0.5\pi x - 4\pi t - {\pi \over 4}} \right)

Compare with a standard wave equation,

y=asin⁡(2πxλ−2πtT+ϕ)y = a\sin \left( {{{2\pi x} \over \lambda } - {{2\pi t} \over T} + \phi } \right)

we get

2πλ=0.5π⇒λ=2π0.5π=4m{{2\pi } \over \lambda } = 0.5\pi \Rightarrow \lambda = {{2\pi } \over {0.5\pi }} = 4m
2πT=4π⇒T=2π4π=12sec⁡{{2\pi } \over T} = 4\pi \Rightarrow T = {{2\pi } \over {4\pi }} = {1 \over 2}\sec
υ=1/T=2 Hz\upsilon = 1/T = 2\,Hz

. Wave velocity,

v=λυ=4×2=8v = \lambda \upsilon = 4 \times 2 = 8

m/sec.

Q50
A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at distances of 2 m and 3 m respectively from the source. The ratio of the intensities of the waves at P and Q is
A 3 : 2
B 2 : 3
C 9 : 4
D 4 : 9
Correct Answer
Option C
Solution

d 1 = 2 m, d 2 = 3 m

Intensity∝1(distance)2Intensity \propto {1 \over {{{\left( {distance} \right)}^2}}}
I1∝122{I_1} \propto {1 \over {{2^2}}}

and

I2∝132{I_2} \propto {1 \over {{3^2}}}
∴I1I2=94\therefore {{{I_1}} \over {{I_2}}} = {9 \over 4}
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