Question

A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.

Answer 1

It is given that:
Amplitude of simple harmonic motion, x = 0.1 m
Time period of simple harmonic motion, T = 0.314 s
Mass of the block, m = 0.5 kg
Weight of the block, W = mg = 0.5$\times$10 = 5 kg $\left(\because g=10{\mathrm{ms}}^{-2}\right)$

Total force exerted on the block = Weight of the block + spring force



Periodic time of spring is given by,
$$\begin{gathered} \mathrm{T}=2\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)} \\ \Rightarrow 0.314=2\mathrm{\pi }\sqrt{\left(\frac{0.5}{k}\right)} \\ \Rightarrow k=200\mathrm{N}/\mathrm{m} \end{gathered}$$

∴ The force exerted by the spring on the block $\left(F\right)$ is,
F = kx = 200.0 × 0.1 = 20 N

Maximum force = F + weight of the block
= 20 + 5 = 25 N