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प्रश्न
A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4 s. If the oscillations are stopped and the body hangs in equilibrium find the potential energy stored in the spring.
उत्तर
It is given that:
Mass of the body, m = 2 kg
Time period of the spring mass system, T = 4 s
The time period for spring-mass system is given as,
\[T = 2\pi\sqrt{\left( \frac{m}{k} \right)}\]
where k is the spring constant.
On substituting the respective values, we get:
\[\Rightarrow 4 = 2\pi\sqrt{\frac{2}{k}}\]
\[ \Rightarrow 2 = \pi\sqrt{\frac{2}{k}}\]
\[ \Rightarrow 4 = \pi^2 \left( \frac{2}{k} \right)\]
\[ \Rightarrow k = \frac{2 \pi^2}{4}\]
\[ = \frac{\pi^2}{2} = 5 N/m\]
As the restoring force is balanced by the weight, we can write:
F = mg = kx
\[\Rightarrow x = \frac{mg}{k} = \frac{2 \times 10}{5} = 4\]
∴ Potential Energy \[\left( U \right)\] of the spring is,
\[U = \left( \frac{1}{2} \right)k x^2 = \frac{1}{2} \times 5 \times 16\]
\[ = 5 \times 8 = 40 J\]
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