First slide

Simple harmonic motion

Question

A small body of mass 0.10 kg is executing S.H.M. of amplitude 1.0 m and period 0.20 sec. The maximum force acting on it is

Moderate

A

98.596 N
B

985.96 N
C

100.2 N
D

76.23 N

Solution

Maximum acceleration
$A_{\max}$ $= {aω}^{2} = \frac{a \times 4 π^{2}}{T^{2}}$ $= \frac{1 \times 4 \times {(3 .14)}^{2}}{0 .2 \times 0 .2}$
$F_{\max} = m \times A_{\max} = \frac{0 .1 \times 4 \times {(3 .14)}^{2}}{0 .2 \times 0 .2} = 98 .596 N$

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