Question

The displacement (in m) of a particle of mass 100 gram from its equilibrium position is given by $y=0.01\sin 2\pi (t+0.4)$. Select the correct option(s) :

• A. The time period of motion is 1 sec.
• B. The time period of motion is $\frac{1}{7.5}sec$
• C. The maximum acceleration of the particle is $0.04{\pi }^{2}m/s^2$
• D. The force acting on the particle is zero when the displacement is 0.05m

Answer 1

Answer: (A), (C)

The correct options are
A The time period of motion is 1 sec.
C The maximum acceleration of the particle is $0.04{\pi }^{2}m/s^2$

$\omega =2\pi s^{-1}T=\frac{2\pi }{\omega }=\frac{2\pi }{2\pi }=1s$

So 'A' is correct and 'B' is incorrect.
$a_{max}={\omega }^{2}Am/s^2=(2\pi {)}^{2}Am/s^2=0.04{\pi }^{2}m/s^2$
So option 'C' is correct.
And $0.05m$ displacement is not possible. Since its maximum Displacement is $0.01m$. So we can say that option 'D' is also incorrect Since it will never attain a displacement of $0.05m$.