Question

A body of mass $0.5$ kg is performing S.H.M. with a time period $\pi /2$ seconds. If its velocity at mean position is $1$ m/s, the restoring force acts on the body at a phase angle ${60}^o$ from extreme position is

• A. $0.5N$
• B. $1N$
• C. $2N$
• D. $4N$

Answer 1

The correct option is B $1N$

$T=\frac{\pi }{2}$
$V_{max}=1m/sec$
$\omega =\frac{2\pi }{T}$
$V=4rad/sec$
$V_{max=}A\omega$
$A\times 4=1$
$A=\frac{1}{4}$
$a={\omega }^{2}x$
$F=m{\omega }^{2}x$
$x=Acos{60}^o$
$\therefore x=\frac{A}{2}$
$F=0.5\times (4{)}^2\times \frac{1}{4}\times \frac{1}{2}$
$F=1N$