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 $=x_{\max }=\sqrt{3^{2}+1^{2}}=\sqrt{10}, \omega=5 \pi$
$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{5 \pi}=0.4 s$