First slide

Simple hormonic motion

Question

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 $m / s^{2}$ at a distance of 5 m from the mean position. The time period of oscillation is

Moderate

A

l s
B

2 $π$ s
C

2 s
D

$π s$

Solution

In case of SHM $\vec{a} = - ω^{2} \vec{y}$
Given a = 20 $m / s^{2}$ at y = 5 m
Hence $| \vec{a} | = ω^{2} | \vec{y} |$
$20 = ω^{2} \times 5$
or $ω^{2} = 4 \Rightarrow ω = 2 rad / \sec$
But $ω = \frac{2 π}{T}$
$2 = \frac{2 π}{T} o r T = π \sec$

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