Question

6) When the displacement of a simple harmonic oscillator is half of its amplitude, its
P.E. is 3 J. Its total energy is
a) 6J
b) 12 J c) 15 J d) 20 J​

Answer 1

Answer:

b) 12 J

Explanation:

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Answer 2

Given:

The potential energy of a simple harmonic oscillator is 3J when displaced by half of its amplitude.

To Find:

The total energy of the simple harmonic oscillator.

Solution:

We know that the potential energy of a simple harmonic oscillator is given by $PE=\frac{1}{2} kx^2$ where $k$ is the force constant and $x$ is the displacement of the oscillator from its mean position.

At the extreme position, since the particle comes into rest, so there is no kinetic energy and the potential energy $\frac{1}{2} kA^2$ becomes equal to the total energy, where $A$ is the amplitude of the oscillator.

We're given that the total potential energy when the oscillator is displaced by half of its amplitude $\frac{A}{2}$ is equal to 3J.

⇒ $\frac{1}{2}k(\frac{A}{2} )^2=3J$

⇒ $\frac{1}{4}* \frac{1}{2}k(A )^2=3J$

⇒ $\frac{1}{2}kA ^2=12J$

Hence, the total energy of the simple harmonic oscillator is equal to 12J.