A body of mass 2 kg suspended from a vertical spring undergoes simple harmonic motion with a period of 4 s . If the oscillations are halted and the body hangs in equilibrium, determine the potential energy stored in the

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Quantrex Academy · Accepted Answer

Correct answer: D. 40 J

The time period of the spring-mass system is given by

T = 2\pi \sqrt{\dfrac{m}{k}}

.

Substituting the given values:

4 = 2\pi \sqrt{\dfrac{2}{k}}

Squaring both sides:

16 = 4\pi^2 \left(\dfrac{2}{k}\right)<br><br>

k =

\dfrac{8\pi^2}{16} = \dfrac{\pi^2}{2} \text{ N/m}<br><br>Taking

\pi^2 \approx 10

, we get

k = 5 \text{ N/m}

.<br><br>At equilibrium, the spring force balances the weight of the body:<br><br>

kx = mg

<br><br>

x =

\dfrac{mg}{k} = \dfrac{2 \times 10}{5} =

4 \text{ m}

<br><br>The potential energy stored in the spring at equilibrium is:<br><br>

U =

\dfrac{1}{2}

kx^2

<br><br>

U =

\dfrac{1}{2} \times

5 \times (4)^2 = 40 \text{ J}

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