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Waves

Section: Physics

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Question : 31 of 31

Marks: +1, -0

A transverse sinusoidal wave moves along a string in the positive x-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 t, the snapshot of the wave is shown in the figure. The velocity of point P when its displacement is 5 cm is

[JEE Adv 2008 P2]

$\frac{\sqrt{3}\pi}{50} \hat{j}$ m/s
- $\frac{\sqrt{3}\pi}{50} \hat{j}$ m/s
$\frac{\sqrt{3}\pi}{50} \hat{i}$ m/s
- $\frac{\sqrt{3}\pi}{50} \hat{i}$ m/s

Solution:

From the figure, the phase of the wave is zero; hence, the equation of the wave is

y = Asinωt

where A is the amplitude (A = 10 cm) and w is the angular frequency.

v = Aω cosωt

It is given that y = 5 cm; thus,

5 = 10 sinωθ ⇒ ωθ = 30°

Therefore, the velocity is

v = 10 cm × 2π ×

\frac{10}{50\,\mathrm{cm}}

cos 30°

=

\frac{200\pi}{100\times 5} \times \frac{\sqrt{3}}{2}

=

\frac{\pi\sqrt{3}}{50} \hat{j}

m/s

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