Question

A body of mass 0.8 kg performs linear S.H.M. It experiences a restoring force of 0.4N, when its displacement from mean position is 4 cm. Determine Force constant and Period of S.H.M.

Hint:

S.H.M. problems often have two steps: First, use given force and displacement to find the spring/force constant (\(k\)). Second, use \(k\) and the mass to find the period or frequency.

Answer 1

Given:
Mass of the body, \(m = 0.8 \, kg\)
Restoring force, \(F = 0.4 \, N\)
Displacement, \(x = 4 \, cm = 0.04 \, m\)
1. To find the Force Constant (k):
The restoring force in Simple Harmonic Motion (S.H.M.) is given by Hooke's Law, \(F = kx\) (considering magnitude). \[ k = \frac{F}{x} \] \[ k = \frac{0.4 \, N}{0.04 \, m} = 10 \, N/m \] The force constant is 10 N/m.
2. To find the Period of S.H.M. (T):
The period of S.H.M. is given by the formula: \[ T = 2\pi\sqrt{\frac{m}{k}} \] \[ T = 2\pi\sqrt{\frac{0.8}{10}} = 2\pi\sqrt{0.08} = 2\pi\sqrt{\frac{8}{100}} \] \[ T = 2\pi\frac{\sqrt{8}}{10} = 2\pi\frac{2\sqrt{2}}{10} = \frac{2\sqrt{2}\pi}{5} \, s \] Using \(\pi \approx 3.14\) and \(\sqrt{2} \approx 1.414\): \[ T \approx \frac{2 \times 1.414 \times 3.14}{5} \approx \frac{8.88}{5} \approx 1.776 \, s \] The period of S.H.M. is approximately 1.78 s.