Question

The displacement ( in m) of a particle of mass 100 g from its equilibrium position is given by the equation:
$y=0.05\sin 3\pi (5t+0.4)$

• A. the time period of motion is $1/30$ sec
• B. the time period of motion is $1/705$ sec
• C. the maximum acceleration of the particle is $11.25{\pi }^{2}m/s^2$
• D. the force acting on the particle is zero when the displacement is 0.05 m.

Answer 1

The correct option is C the maximum acceleration of the particle is $11.25{\pi }^{2}m/s^2$

$\begin{array}{c}y=0.05\sin \left(15\pi t+1.2\pi \right)\\ w=15\pi =\frac{2\pi }{T}\\ T=\frac{2}{15}\\ a_{max}=w^{2}A=225{\pi }^2\times 0.05\\ =11.25{\pi }^{2}m/s^2\\ Hence,theoptionCisthecorrectanswer.\end{array}$