Question 70: Oscillations

Question

A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

Accepted Answer

Let's denote the amplitude of the simple harmonic motion as A, and the displacement of the particle from the mean position as x. The given condition is that x = A/2.
For a particle in SHM, the potential energy (PE) is given by:
$$PE = \frac{1}{2} kx^2$$
And the total mechanical energy (E) of the particle remains constant and is given by:
$$E = \frac{1}{2} kA^2$$
Since the total mechanical energy is the sum of potential energy and kinetic energy (KE), we have:
$$E = PE + KE$$
Now, we need to find the ratio of potential energy to kinetic energy when x = A/2.
Calculate the potential energy at x = A/2:
$$PE = \frac{1}{2} k\left(\frac{A}{2}\right)^2 = \frac{1}{8} kA^2$$
Substitute the expression for total mechanical energy:
$$KE = E - PE = \frac{1}{2} kA^2 - \frac{1}{8} kA^2 = \frac{3}{8} kA^2$$
Now, find the ratio of potential energy to kinetic energy:
$$\frac{PE}{KE} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}$$
Therefore, the ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude is 1 : 3.

Answer

1 : 1

Answer

1 : 4

Answer

2 : 1

Answer

1 : 3

Oscillations

A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

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