Question

A particle of mass 10g is executing simple harmonic motion with an amplitude of 0.5m and periodic time of $\frac{\pi }{5}s$. The maximum value of the force acting on the particle is

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

Answer 1

The correct option is D 0.5 N.

Given,

Mass = m = 10 g

Amplitude = A = 0.5 m

Time period = T = $\frac{\pi }{5}s$

From, $F=ma$ we know that for constant $m$ the force experienced by the body will be maximum when acceleration $a$ is maximum.

For a particle executing simple harmonic motion the acceleration at a displacement $x$ is given by,

$a=-{\omega }^{2}x$

where $\omega =\frac{2\pi }{T}$ is the angular frequency of the SHM.

Hence acceleration will be maximum at the maximum displacement i.e.; at x = A.

$\therefore F_{max}=m{\omega }^{2}A=m\frac{4{\pi }^2}{T^2}A$
$F_{max}=\frac{10}{1000}\times \frac{4\times (\pi {)}^2}{(\frac{\pi }{5}{)}^2}\times 0.5=0.5N$.