When a body is executing Simple Harmonic Motion (SHM), its velocity $$v$$ at a displacement $$x$$ from the mean position can be given by the equation:

$$v = \omega \sqrt{A^2 - x^2}$$

Here, $$\omega$$ is the angular frequency and $$A$$ is the amplitude of the SHM.

Given:

At displacement $$4 \mathrm{~cm}$$, velocity $$10 \mathrm{~cms}^{-1}$$
At displacement $$5 \mathrm{~cm}$$, velocity $$8 \mathrm{~cms}^{-1}$$

We use the equation for velocity at displacement for both given conditions to form two equations:

At $$x = 4 \mathrm{~cm}$$ and $$v = 10 \mathrm{~cms}^{-1}$$:

$$10 = \omega \sqrt{A^2 - 4^2}$$

This can be rewritten as:

$$10 = \omega \sqrt{A^2 - 16}$$

At $$x = 5 \mathrm{~cm}$$ and $$v = 8 \mathrm{~cms}^{-1}$$:

$$8 = \omega \sqrt{A^2 - 5^2}$$

This can be rewritten as:

$$8 = \omega \sqrt{A^2 - 25}$$

Now, we square both equations and equate them to form a system of equations:

For the first equation:

$$100 = \omega^2 (A^2 - 16)$$

For the second equation:

$$64 = \omega^2 (A^2 - 25)$$

Divide the first equation by the second equation to eliminate $$\omega^2$$:

$$\frac{100}{64} = \frac{A^2 - 16}{A^2 - 25}$$

Simplify:

$$\frac{25}{16} = \frac{A^2 - 16}{A^2 - 25}$$

Cross-multiplying gives:

$$25(A^2 - 25) = 16(A^2 - 16)$$

Expand and simplify:

$$25A^2 - 625 = 16A^2 - 256$$

Rearrange terms:

$$9A^2 = 369$$

Therefore:

$$A^2 = 41$$

The amplitude $$A = \sqrt{41}$$.

We can now use either set of velocity and displacement to find $$\omega$$. Using the first set:

$$10 = \omega \sqrt{41 - 16}$$

$$10 = \omega \sqrt{25}$$

Therefore:

$$10 = 5\omega$$

$$\omega = 2 \mathrm{~rad/s}$$

The periodic time $$T$$ is given by the relation:

$$T = \frac{2\pi}{\omega}$$

Substitute $$\omega$$:

$$T = \frac{2\pi}{2} = \pi \mathrm{~sec}$$

Therefore, the periodic time $$\mathrm{t}$$ is Option D: $$\pi$$ sec.

Simple Harmonic Motion

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