First slide

Simple harmonic motion

Question

A particle executing simple harmonic motion has an amplitude of 6 cm. Its acceleration at a distance of 2 cm from the mean position is $8 cm / s^{2}$ . The maximum speed of the particle is

Moderate

A

8 cm/s
B

12 cm/s
C

16 cm/s
D

24 cm/s

Solution

$A = ω^{2} y$ $\Rightarrow ω = \sqrt{A / y} = \sqrt{\frac{8}{2}} = 2 rad / \sec$
Now $v_{\max} = aω = 6 \times 2 = 12 cm / \sec$

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