Question

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

• A. $98.596N$
• B. $985.96N$
• C. $100.2N$
• D. $76.23N$

Answer 1

Answer: (A)

$98.596N$

Given:

The mass of the body is $m=0.10kg$

The amplitude of SHM $a=1.0m$.

The time period of oscillation is $T=0.20s$

Maximum acceleration of the syustem is given by:

$A_{max}=a{\omega }^2$ $\because \omega =\frac{2\pi }{T}$

$=\frac{a\times 4{\pi }^2}{T^2}=\frac{1\times 4\times (3.14{)}^2}{0.2\times 0.2}$

So, the maximum force acting on the body will be:

$F_{max}=m\times A_{max}=\frac{0.1\times 4\times (3.14{)}^2}{0.2\times 0.2}=98.596N$