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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.
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Solution
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
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