First slide

Simple harmonic motion

Question

A body executing SHM has a maximum acceleration equal to $48 m / s e c^{2}$ and maximum velocity equal to 12 m/sec. The amplitude of SHM is

Moderate

A

3 m
B

$\frac{3}{32} m$
C

$\frac{1024}{9} m$
D

$\frac{64}{9} m$

Solution

given maximum acceleration $a_{max} {=48m/s}^{2}$
Maximum velocity $V_{max} =12m/s$
amplitude A=? here angular velocity = $ω$
$\frac{a_{max}}{V_{max}} = \frac{ω^{2} A}{ωA} = \frac{48}{12} = 4$
=> $ω$ =4
$V_{max} = ωA$
substitute maximum velocity, and angular velocity
12=4(A)
A=3m.

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