Question

A loaded vertical spring executes S.H.M. with a time period of 4 sec. The difference between the kinetic energy and potential energy of this system varies with a period of

• A. 2 sec
• B. 1 sec
• C. 8 sec
• D. 4 sec

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between the kinetic energy (KE) and potential energy (PE) in a system undergoing Simple Harmonic Motion (SHM).

The time period of the system T = 4 \, \text{sec}. In SHM, the time period is the time taken for one complete cycle of motion.

The kinetic energy (KE) and potential energy (PE) of the system vary such that:

• When the displacement is maximum (at the amplitude), the potential energy is maximum, and kinetic energy is zero.
• When the displacement is zero (at the mean position), the kinetic energy is maximum, and potential energy is zero.

Both KE and PE vary with a period equal to the time period of the system, which is 4 seconds.

However, the problem asks us about the period with which the difference between the kinetic energy and potential energy (KE - PE) varies. To deduce this, consider that:

• The energy difference itself is a periodic function. We will consider its mathematical form.

In SHM, at any time t, the sum of kinetic and potential energy is constant, equal to the total mechanical energy:

E = KE + PE

The difference (KE - PE) is also a periodic function. Notably, because the energies oscillate, this difference reaches its full cycle twice in one period of motion. The reason is that the energies switch roles every half cycle, i.e., maximum kinetic energy corresponds to zero potential energy and vice versa. This consideration leads us to the conclusion that the period of variation of the energy difference is half of the SHM period.

Therefore, the period during which the difference between kinetic and potential energies varies is:

\frac{T}{2} = \frac{4 \, \text{sec}}{2} = 2 \, \text{sec}

Thus, the correct answer is 2 sec.