A spring of spring constant 2000N/m is suspended vertically in earth's gravitational field and a mass of 0.8 kg is hung from it. It is hanging in equilibrium. At t = 0, a force of 72 N starts acting in vertically downward direction. The force ceases to act at t = \frac{\pi}{2}sec. Find the amplitude of resulting SHM (in mm) after that.

Engineering

Physics

Simple Harmonic Motion

Question

A spring of spring constant 2000N/m is suspended vertically in earth's gravitational field and a mass of 0.8 kg is hung from it. It is hanging in equilibrium. At t = 0, a force of 72 N starts acting in vertically downward direction. The force ceases to act at t = $\frac{π}{2}$ sec. Find the amplitude of resulting SHM (in mm) after that.

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At O, Mg = ky₀

${OO}^{'} = \frac{F}{k} and O^{'} A = \frac{F}{k}$

$T = 2 π \sqrt{\frac{m}{k}} = \frac{π}{25} sec.$

$at t = \frac{π}{2} sec$

number of oscillations $N = \frac{π / 2}{π / 25}$

N = 12.5 ; it means when force cases to act, body is at position A.

So amplitude of resulting SHM, $OA = \frac{2 F}{k} = \frac{2 \times 72}{2000} = 72 mm$

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