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

$x = r = 0.1 m T = 0.314 s e c, m = 0.5 k g$

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

$T = 2 π \sqrt{(\frac{m}{k})}$

$\Rightarrow 0.314 = 2 π \sqrt{(\frac{0.5}{k})}$

$\Rightarrow k = 200 N / m$

$∴$ Force exerted by the spring on the block is,

$F = k x = 201.1 \times 01 = 20 N$

$∴$ Maximum force = F + weight = 20 + 5 = 25 N

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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.

x=r=0.1 m T=0.314 sec, m=0.5 kg Total force exerted on the block = Weight of the block + spring force T=2π √(mk) ⇒ 0.314=2π √(0.5k) ⇒ k=200 N/m ∴ Force exerted by the spring on the block is, F=kx=201.1×01=20 N ∴ Maximum force = F + weight = 20 + 5 = 25 N