A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is:

We are told that the motion is simple harmonic, so the particle’s displacement from the mean position can be written as $$x(t)=A\sin(\omega t+\phi)$$. For any SHM the following standard relations hold:

1. The instantaneous speed (magnitude of velocity) is given by the well-known formula $$v=\omega\sqrt{A^{2}-x^{2}}.$$ 2. The instantaneous acceleration (its magnitude) is obtained from the second derivative of displacement and is $$a=\omega^{2}x.$$

The data given in the question are

Amplitude $$A = 5\ \text{cm} = 0.05\ \text{m},$$ Instantaneous displacement $$x = 4\ \text{cm} = 0.04\ \text{m}.$$

The statement “the magnitude of its velocity is equal to the magnitude of its acceleration” means that, numerically in SI units,

$$v = a.$$

Substituting the standard expressions for $$v$$ and $$a$$ we get

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

The angular frequency $$\omega$$ is not zero, so we can divide both sides of the above equation by $$\omega$$ to obtain

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

Now we substitute the given values of $$A$$ and $$x$$ (we may keep them in centimetres because the ratio will be dimensionless; the numeric equality is all that matters):

First calculate $$A^{2}-x^{2}$$: $$A^{2}-x^{2}=5^{2}-4^{2}=25-16=9.$$

Therefore $$\sqrt{A^{2}-x^{2}}=\sqrt{9}=3\ \text{cm}.$$

Putting this into the earlier relation gives

$$3=\omega\,(4).$$

Hence the angular frequency is

$$\omega=\frac{3}{4}\ \text{s}^{-1}.$$

The periodic time $$T$$ (time‐period) of an SHM is connected to the angular frequency by the basic relation

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

Substituting $$\omega=\dfrac{3}{4}\ \text{s}^{-1}$$ we get

$$T=\frac{2\pi}{3/4}=2\pi\times\frac{4}{3}=\frac{8\pi}{3}\ \text{s}.$$

Hence, the correct answer is Option A.