Question

A body of mass 1/4 kg is in S.H.M and its displacement is given by the relation $y=0.05sin(20t+\frac{\pi }{2})$ m. If $t$ is in seconds, the maximum force acting on the particle is:

• A. $5$ N
• B. $2.5$ N
• C. $10$ N
• D. $0.25$ N

Answer 1

Answer: (A)

$5$ N

We know that

$F=m{\omega }^{2}A$
$\omega =20rad/sec$
$A=0.05m$
Thus
$F=\frac{1}{4}\times 20\times 20\times \frac{1}{20}$
$=5N$