A particle of mass 1 kg is hanging from a spring of force constant 100 N m$$^{-1}$$. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period $$T$$. The time when the kinetic energy and potential energy of the system will become equal, is $$\frac{T}{n}$$. The value of $$n$$ is _________.

We have a block-spring system in which the mass is $$m = 1\ \text{kg}$$ and the spring constant is $$k = 100\ \text{N\,m}^{-1}$$. When the mass is pulled slightly and released it performs free simple harmonic motion (SHM).

First recall the angular frequency-period relation for SHM:

$$\omega = \sqrt{\dfrac{k}{m}} , \quad T = \dfrac{2\pi}{\omega}.$$

Substituting the given values,

$$\omega = \sqrt{\dfrac{100}{1}} = 10\ \text{rad s}^{-1},$$

and then

$$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{10} = \dfrac{\pi}{5}\ \text{s}.$$

For SHM the displacement as a function of time may be written as

$$x = A\sin(\omega t + \phi),$$

where $$A$$ is the amplitude. We can choose the phase constant $$\phi = 0$$ by measuring time from the instant when the mass is at the extreme position. Thus,

$$x = A\sin\omega t.$$

The potential energy (elastic energy) stored in the spring at any instant is

$$U = \dfrac12 kx^{2} = \dfrac12 kA^{2}\sin^{2}\omega t.$$

The kinetic energy of the mass is

$$K = \dfrac12 mv^{2}, \quad\text{where}\quad v = \dfrac{dx}{dt} = A\omega\cos\omega t.$$

Hence,

$$K = \dfrac12 mA^{2}\omega^{2}\cos^{2}\omega t.$$

But for a mass-spring system $$m\omega^{2} = k,$$ so we may replace $$m\omega^{2}$$ by $$k$$:

$$K = \dfrac12 kA^{2}\cos^{2}\omega t.$$

Now we set the kinetic and potential energies equal:

$$K = U \;\;\Longrightarrow\;\; \dfrac12 kA^{2}\cos^{2}\omega t = \dfrac12 kA^{2}\sin^{2}\omega t.$$

Cancelling the common non-zero factors $$\dfrac12 kA^{2},$$ we get

$$\cos^{2}\omega t = \sin^{2}\omega t.$$

Taking square roots (and noting that both energies are positive),

$$\cos\omega t = \pm \sin\omega t.$$

This can be rewritten as

$$\tan\omega t = \pm 1.$$

The smallest positive solution for $$\omega t$$ satisfying $$\tan\omega t = 1$$ is

$$\omega t = \dfrac{\pi}{4}.$$

Solving for the time $$t$$ gives

$$t = \dfrac{\pi/4}{\omega}.$$

But $$\omega = \dfrac{2\pi}{T},$$ so

$$t = \dfrac{\pi/4}{2\pi/T} = \dfrac{\pi}{4}\cdot\dfrac{T}{2\pi} = \dfrac{T}{8}.$$

Thus the first instant when the kinetic energy equals the potential energy occurs after a time $$\dfrac{T}{8}$$ from the start of the motion. Comparing with the form $$\dfrac{T}{n},$$ we identify $$n = 8.$$

Hence, the correct answer is Option 8.