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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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A particle on the trough of a wave at any instant will come to the mean position after a time (T = time period)

T/2

T/4

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Explanation

The particle will come after a time $\frac{T}{4}$ to its mean position.

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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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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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