Physics - SHM Question with Solution | TestHub

Given time period $T=16$ s.

Angular frequency $\omega =\frac{2\pi }{T}=\frac{2\pi }{16}=\frac{\pi }{8}$ rad/s.

Since the particle crosses the mean position at $t=2$ s, the displacement equation can be written as:

$x=A\sin (\omega (t-2))$

The velocity of the particle is given by the derivative of displacement with respect to time:

$v=\frac{dx}{dt}=A\omega \cos (\omega (t-2))$

Given that at $t=4$ s, the velocity is $v=4$ m/s:

$4=A\left(\frac{\pi }{8}\right)\cos \left(\frac{\pi }{8}(4-2)\right)$

$4=A\left(\frac{\pi }{8}\right)\cos \left(\frac{\pi }{4}\right)$

Substituting $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$:

$4=A\left(\frac{\pi }{8}\right)\left(\frac{1}{\sqrt{2}}\right)$

$A=\frac{32\sqrt{2}}{\pi }$ m

Answer: $32\sqrt{2}/\pi$