First slide

Simple hormonic motion

Question

A particle is executing simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm/sec. The distance of the particle from the mean position when its speed becomes 5 cm/s is

Moderate

A

$\sqrt{3} cm$
B

$2 \sqrt{2} cm$
C

$2 \sqrt{3} cm$
D

$3 \sqrt{2} cm$

Solution

$v = ω \sqrt{a^{2} - y^{2}}$
At y = 0, $v_{\max} = 10 cm / s$
As a = 4 cm, $v_{\max} = ωa hence ω = 2.5 rad / s$
Hence, $5 = 2.5 \sqrt{16 - y^{2}}$
or y = $2 \sqrt{3} cm$

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