Single Choice

A particle executes SHM of period $$12$$s. After two seconds, it passes through the centre of oscillation, the velocity is found to be $$3.142 cm s^{-1}$$. The amplitude of oscillation is:

A$$6$$cm
B$$3$$ cm
C$$24$$cm
D$$12$$cm
Correct Answer

Solution

Let $$y = A sin \omega t$$ Then $$v = \dfrac{dy}{dt} = \omega A cos \omega t = \dfrac{2 \pi}{T} A cos \dfrac{2 \pi}{T}t$$ $$\therefore 3.142 = \dfrac{2 \times 3.142}{12} \times A cos \dfrac{2 \pi}{12} \times 2$$ $$\therefore A = 12cm$$

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