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Waves MCQs for NEET — Physics Questions with Answers

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When two sound waves with a phase difference of π/2, and each having amplitude A and frequency ω, are superimposed on each other, then the maximum amplitude and frequency of the resultant wave is :

$\frac{A}{\sqrt{2}} : \frac{ω}{2}$

$\frac{A}{\sqrt{2}} : ω$

$\sqrt{2} A : \frac{ω}{2}$

$\sqrt{2} A : ω$

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Explanation

$A_{\max} = \sqrt{A^{2} + A^{2}} = A \sqrt{2},$ frequency will remain same i.e. ω.

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Two waves are propagating to the point P along a straight line produced by two sources A and B of simple harmonic and of equal frequency. The amplitude of every wave at P is ‘a’ and the phase of A is ahead by π/3 than that of B and the distance AP is greater than BP by 50 cm. Then the resultant amplitude at the point P will be, if the wavelength is 1 meter is -

$a \sqrt{3}$

$a \sqrt{2}$

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Explanation

Path difference $(Δ x) = 50 c m = \frac{1}{2} m$

∴ Phase difference $Δ φ = \frac{2 π}{λ} \times Δ x \Rightarrow φ = \frac{2 π}{1} \times \frac{1}{2} = π$

Total phase difference = $π - \frac{π}{3} = \frac{2 π}{3}$

⇒ $A = \sqrt{a^{2} + a^{2} + 2 a^{2} \cos (2 π / 3)} = a$

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The minimum intensity of sound is zero at a point due to two sources of nearly equal frequencies, when :

Two sources are vibrating in opposite phase

The amplitude of the two sources are equal

At the point of observation, the amplitudes of two S.H.M. produced by two sources are equal and both the S.H.M. are along the same straight line

Both the sources are in the same phase

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Explanation

If two waves of nearly equal frequency superpose, they give beats if they both travel in straight line and I_min = 0 if they have equal amplitudes.

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Two sound waves (expressed in CGS units) given by $y_{1} = 0 .3 \sin \frac{2 π}{λ} (v t - x)$ and $y_{2} = 0 .4 \sin \frac{2 π}{λ} (v t - x + θ)$ interfere. The resultant amplitude at a place where the phase difference is π/2 will be :

0.7 cm

0.1 cm

0.5 cm

$\frac{1}{10} \sqrt{7} c m$

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Explanation

Resultant amplitude = $\sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos ϕ}$

= $\sqrt{0 {.3}^{2} + 0 {.4}^{2} + 2 \times 0 .3 \times 0 .4 \times \cos \frac{π}{2}}$ = $0 .5 c m$

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If two waves having amplitudes 2A and A and same frequency and velocity, propagate in the same direction in the same phase, the resulting amplitude will be

$\sqrt{5} A$

$\sqrt{2} A$

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Explanation

In the same phase φ = 0 so resultant amplitude = $a_{1} + a_{2} = 2 A + A = 3 A$

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The intensity ratio of the two waves is 1 : 16. The ratio of their amplitudes is

1 : 16

1 : 4

4 : 1

2 : 1

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Explanation

$\frac{I_{1}}{I_{2}} = {(\frac{a_{1}}{a_{2}})}^{2} = \frac{1}{16} \Rightarrow \frac{a_{1}}{a_{2}} = \frac{1}{4}$

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The superposing waves are represented by the following equations : $y_{1} = 5 \sin 2 π (10 t - 0 .1 x)$ , $y_{2} = 10 \sin 2 π (20 t - 0 .2 x)$ Ratio of intensities $\frac{I_{\max}}{I_{\min}}$ will be :

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Explanation

$a_{1} = 5, a_{2} = 10$

$\Rightarrow \frac{I_{\max}}{I_{\min}} = \frac{{(a_{1} + a_{2})}^{2}}{{(a_{1} - a_{2})}^{2}} = {(\frac{5 + 10}{5 - 10})}^{2} = \frac{9}{1}$

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The two interfering waves have intensities in the ratio 9 : 4. The ratio of intensities of maxima and minima in the interference pattern will be :

1 : 25

25 : 1

9 : 4

4 : 9

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Explanation

$\frac{I_{\max}}{I_{\min}} = {(\frac{\sqrt{\frac{I_{1}}{I_{2}}} + 1}{\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} - 1})}^{2} = {(\frac{\sqrt{\frac{9}{4}} + 1}{\sqrt{\frac{9}{4}} - 2})}^{2} = \frac{25}{1}$

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If the ratio of amplitude of two waves is 4 : 3. Then the ratio of maximum and minimum intensity will be :

16 : 18

18 : 16

49 : 1

1 : 49

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Explanation

$\frac{I_{\max}}{I_{\min}} = {(\frac{\frac{a_{1}}{a_{2}} + 1}{\frac{a_{1}}{a_{2}} - 1})}^{2} = {(\frac{\frac{4}{3} + 1}{\frac{4}{3} - 1})}^{2} = \frac{49}{1}$

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Equation of motion in the same direction is given by $y_{1} = A \sin (ω t - k x)$ , $y_{2} = A \sin (ω t - k x - θ)$ . The amplitude of the medium particle will be

$2 A \cos \frac{θ}{2}$

$2 A \cos θ$

$\sqrt{2} A \cos \frac{θ}{2}$

$1 .2 f, 1 .2 λ$

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Explanation

The resultant amplitude is given by

$A_{R} = \sqrt{A^{2} + A^{2} + 2 A A \cos θ} = \sqrt{2 A^{2} (1 + \cos θ)}$

$= 2 A \cos θ / 2$ $(∵ H \cos θ = 2 \cos^{2} θ / 2)$

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