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IIT-JEE 2008 · Paper 2, Physics question 25

A transverse sinusoidal wave moves along a string in the positive xx-direction at a speed of 10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time tt, the snap-shot of the wave is shown in figure. The velocity of point P when its displacement is 5 cm is

Snapshot of a sine wave y against x starting at the origin and rising to a crest, then descending; P is a point on the descending part of the first crest, above the x-axis (between the crest and the zero crossing).
Choose one option
Answer and worked solution

Official answer: A

Direction. For a wave moving in +x+x, vy=−v∂y∂xv_y=-v\dfrac{\partial y}{\partial x}. P is on the falling side of the crest, where the slope is negative, so P moves up (+j^+\hat j).

Magnitude. f=vλ=0.10.5=0.2f=\frac v\lambda=\frac{0.1}{0.5}=0.2 Hz, ω=0.4π\omega=0.4\pi rad/s, A=0.1A=0.1 m. At y=0.05y=0.05 m:
vP=ωA2−y2=0.4π0.0075=0.4π×0.0866=3π50 m/s.v_P=\omega\sqrt{A^2-y^2}=0.4\pi\sqrt{0.0075}=0.4\pi\times0.0866=\frac{\sqrt3\pi}{50}\text{ m/s}.

(A)

Topics:Wave equation and progressive waves
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