First slide

Simple hormonic motion

Question

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is l0 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is

Moderate

A

$\sqrt{3} cm$
B

$\sqrt{5} cm$
C

$2 (\sqrt{3}) cm$
D

$2 (\sqrt{5}) cm$

Solution

$v_{\max} = aω \Rightarrow ω = \frac{v_{\max}}{a} = \frac{10}{4}$
Now, $v = ω \sqrt{a^{2} - y^{2}} \Rightarrow v^{2} = ω^{2} (a^{2} - y^{2})$
$\Rightarrow y^{2} = a^{2} - \frac{v^{2}}{ω^{2}}$
$\Rightarrow Y = \frac{A \sqrt{3}}{2} = 2 \sqrt{3} cm$

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