Physics - SHM Question with Solution | TestHub

Given displacement equation is $x=6\sin \left(100t+\frac{\pi }{4}\right)$ cm.

Comparing with the standard equation of SHM $x=A\sin (\omega t+\varphi )$, we get:

Amplitude $A=6$ cm $=0.06$ m

Angular frequency $\omega =100$ rad/s

Maximum velocity of the body is given by $v_{\text{max}}=A\omega$

$v_{\text{max}}=0.06\times 100=6$ m/s

Maximum kinetic energy is $K_{\text{max}}=\frac{1}{2}m{v}_{\text{max}}^2$

Substituting the values, $K_{\text{max}}=\frac{1}{2}\times 1\times (6{)}^2=18$ J.

Answer: $18$ J