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         \[\left( \because g = 10 {ms}^{- 2} \right)\]

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

Periodic time of spring is given by,

\[T = 2\pi\sqrt{\left( \frac{m}{k} \right)}\]

\[ \Rightarrow 0 . 314 = 2\pi\sqrt{\left( \frac{0 . 5}{k} \right)}\]

\[ \Rightarrow k = 200 N/m\]

∴ 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