Question

A particle oscillates simple harmonically with a period of 16 s. Two second after crossing the equilibrium position its velocity becomes 1 m/s. The amplitude is

• A. $\frac{\pi }{4}$ m
• B. $\frac{8\sqrt{2}}{\pi }$ m
• C. $\frac{8}{\pi }$ m
• D. $\frac{4\sqrt{2}}{\pi }$ m

Answer 1

Answer: (B)

$\frac{8\sqrt{2}}{\pi }$ m

$\omega =\frac{2\pi }{T}=\frac{2\pi }{16}=\frac{\pi }{8}$ rad/s
At t=0, particle crosses the mean position. Hence its velocity is maximum. So, velocity as a function of time can be written as
$v=v_{max}\cos \omega t$
or $v=\omega A\cos \omega t$
$\therefore$ $1=\left(\frac{\pi }{8}\right)A\cos \left(\frac{\pi }{8}\right)\left(2\right)=\left(\frac{\pi }{8\sqrt{2}}\right)A$
$\therefore$ $A=\frac{8\sqrt{2}}{\pi }$ m