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Karnataka Board PUCPUC Science Class 11

P 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

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Question

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.

Sum
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Solution

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:
        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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Energy in Simple Harmonic Motion
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Chapter 12: Simple Harmonics Motion - Exercise [Page 252]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 12 Simple Harmonics Motion
Exercise | Q 12 | Page 252

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