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Waves MCQs for NEET — Physics Questions with Answers

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Two waves of frequencies 20 Hz and 30 Hz travel out from a common point. The phase difference between them after 0.6 sec is :

Zero

$\frac{π}{2}$

$\frac{3 π}{4}$

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Explanation

Beats period = $\frac{1}{30 - 20} = 0 .1$ sec

$Δ ϕ = \frac{2 π}{T} Δ . t = \frac{2 π}{0 .1} \times 0 .6 = 2 π \times 6 = 12 π$ or Zero.

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A simple harmonic progressive wave is represented by the equation : $y = 8 \sin 2 π (0 .1 x - 2 t)$ where x and y are in cm and t is in seconds. At any instant the phase difference between two particles separated by 2.0 cm in the x-direction is :

18°

36°

54°

72°

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Explanation

From the given equation $k = 0 .2 π$

⇒ $\frac{2 π}{λ} = 0 .2 π \Rightarrow λ = 10$ cm

$Δ ϕ = \frac{2 π}{λ} Δ x = \frac{2 π}{10} \times 2 = \frac{2 π}{5} = 72^{o}$

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The equation of progressive wave is $y = a \sin (200 t - x)$ . where x is in meter and t is in second. The velocity of wave is :

200 m/sec

100 m/sec

50 m/sec

None of these

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Explanation

comparing the given equation with $y = a \sin (ω t - k x)$

$ω = 200, k = 1$ so $v = \frac{ω}{k} = 200 m / s$

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A wave equation that gives the displacement along y-direction is given by $y = 0 .001 \sin (100 t + x)$ where x and y are in meter and t is time in seconds. This represented a wave :

Of frequency $\frac{100}{π}$ Hz

Of wavelength one metre

Traveling with a velocity of $\frac{50}{π}$ ms^–1 in the positive X-direction

Traveling with a velocity of 100 ms^–1 in the negative X-direction

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Explanation

Compare the given equation with

$y = a \sin (ω t + k x)$ ⇒ $ω = 2 π n = 100$ ⇒ $n = \frac{50}{π} H z$

$k = \frac{2 π}{λ} = 1$ ⇒ $λ = 2 π$ and $v = ω / k = 100 m / s$

Since ‘+’ is given between t terms and x term, so wave is travelling in negative x-direction.

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A wave travelling in positive X-direction with A = 0.2 m has a velocity of 360 m/sec. if λ = 60 m, then the correct expression for the wave is :

$y = 0 .2 \sin [2 π (6 t + \frac{x}{60})]$

$y = 0 .2 \sin [π (6 t + \frac{x}{60})]$

$y = 0 .2 \sin [2 π (6 t - \frac{x}{60})]$

$y = 0 .2 \sin [π (6 t - \frac{x}{60})]$

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Explanation

A wave travelling in positive x-direction may be represented as $y = A \sin \frac{2 π}{λ} (v t - x)$ . On putting values $y = 0 .2 \sin \frac{2 π}{60} (360 t - x)$ ⇒ $y = 0 .2 \sin 2 π (6 t - \frac{x}{60})$

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Two waves represented by the following equations are travelling in the same medium $y_{1} = 5 \sin 2 π (75 t - 0 .25 x)$ , $y_{2} = 10 \sin 2 π (150 t - 0 .50 x)$

The intensity ratio I₁/I₂ of the two waves is :

1 : 2

1 : 4

1 : 8

1 : 16

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Explanation

$I n t e n s i t y I \propto a^{2} ω^{2} \frac{I_{1}}{I_{2}} = \frac{a_{1}^{2} ω_{1}^{2}}{a_{2}^{2} ω_{2}^{2}} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{16}$

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Which of the following is not true for this progressive wave $y = 4 \sin 2 π (\frac{t}{0 .02} - \frac{x}{100})$ where y and x are in cm and t in sec

Its amplitude is 4 cm

Its wavelength is 100 cm

Its frequency is 50 cycles/sec

Its propagation velocity is 50 × 10³ cm/sec

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Explanation

$y = 4 \sin 2 π (\frac{t}{0 .02} - \frac{x}{100})$ .

Comparing this equation with $y = a \sin 2 π (\frac{t}{T} - \frac{x}{λ})$

$v = \frac{Co-efficient of t}{Co-efficient of x} = \frac{1 / 0 .02}{1 / 100}$

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A transverse progressive wave on a stretched string has a velocity of 10 ms^–1 and a frequency of 100 Hz. The phase difference between two particles of the string which are 2.5 cm apart will be :

$\frac{π}{8}$

$\frac{π}{4}$

$\frac{3 π}{8}$

$\frac{π}{2}$

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Explanation

$v = n λ \Rightarrow λ = 10 c m$

Phase difference= $\frac{2 π}{λ} \times$ Path difference= $\frac{2 π}{10} \times 2 .5$ = $\frac{π}{2}$

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The phase difference between two waves represented by $y_{1} = 10^{- 6} \sin [100 t + (x / 50) + 0 .5] m,$ $y_{2} = 10^{- 6} \cos [100 t + (x / 50)] m$ where x is expressed in metres and t is expressed in seconds, is approximately:

1.5 rad

1.07 rad

2.07 rad

0.5 rad

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Explanation

$y_{1} = 10^{- 6} \sin [100 t + (x / 50) + 0 .5]$

$y_{2} = 10^{- 6} \sin [100 t + (\frac{x}{50}) + (\frac{π}{2})]$

Phase difference $ϕ$

$= [100 t + (x / 50) + 1 .57] - [100 t + (x / 50) + 0 .5]$

$= 1 .07$ radians.

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If the equation of the transverse wave is Y = 2sin(kx – 2t), then the maximum particle velocity is :

4 units

2 units

6 units

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Explanation

Maximum particle velocity $= A ω = 2 \times 2 = 4$ units.

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