Question

A particle of mass $10\, g$ is executing simple harmonic motion with an amplitude of $0.5\, m$ and periodic time of $(\pi / 5)$ second. The maximum value of the force acting on the particle is

• A. 25 N
• B. 5 N
• C. 2.5 N
• D. 0.5 N

Answer 1

Answer: (D)

Force in a simple harmonic motion
$F=m f=m\left(-\omega^{2} x\right)=-m \omega^{2} A \sin \omega t$
For maximum value of $F, \sin \omega t=1 $
$\therefore F_{\max }=m \omega^{2} A$
$=m\left(\frac{2 \pi}{T}\right)^{2} A =\frac{m 4 \pi^{2} A}{T^{2}}$
$=\frac{10 \times 10^{-3} \times 4 \pi^{2} \times 0.5}{\pi^{2} / 25}=0.5 \,N$