Given data

• Amplitude: $A = 5\text{ cm} = 0.05\text{ m}$
• Displacement from mean position: $x = 4\text{ cm} = 0.04\text{ m}$
• In SHM, at this position, magnitude of velocity equals magnitude of acceleration: $v = a$

Formulas for SHM

• Speed at displacement $x$: $v = \omega\sqrt{A^2 - x^2}$
• Acceleration magnitude at displacement $x$: $a = \omega^2 x$

Apply the condition $v=a$ $\omega\sqrt{A^2-x^2} = \omega^2 x$

Since $\omega \neq 0$, divide both sides by $\omega$: $\sqrt{A^2-x^2} = \omega x$

Hence, $\omega = \frac{\sqrt{A^2-x^2}}{x}$

Substitute values $A^2-x^2 = (0.05)^2 - (0.04)^2 = 0.0025 - 0.0016 = 0.0009$

$\sqrt{A^2-x^2} = \sqrt{0.0009} = 0.03$

Therefore, $\omega = \frac{0.03}{0.04} = \frac{3}{4}\text{ rad/s}$

Find the time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{3/4} = 2\pi\cdot \frac{4}{3} = \frac{8\pi}{3}\text{ s}$

Match with options $T = \frac{8\pi}{3}\text{ s}$ This corresponds to Option D.

Comparison with stored answer Stored correct answer: D

Our derived answer also gives D, so they agree.