Question

A particle is executing a motion in which its displacement as a function of time is given by $x=3\sin (5\pi t+\pi /3)+\cos (5\pi t+\pi /3)$ where $x$ is in $m$ and is in s. Then the motion is

• A. simple harmonic with time period 0.2 s
• B. simple harmonic with time period 0.4 s
• C. simple harmonic with amplitude 3 m
• D. not a simple harmonic but a periodic motion

Answer 1

Answer: (B)

$x=3\sin (5\pi t+\pi /3)+\cos (5\pi t+\pi /3)$
Amplitude $=x_{\max }=\sqrt{3^2+1^2}=\sqrt{10},\omega =5\pi$
$T=\frac{2\pi }{\omega }=\frac{2\pi }{5\pi }=0.4s$