A particle starts with S.H.M. from the mean position as shown in the figure. Its amplitude is A and its time period is T. At one time, its speed is half that of the maximum speed. What is its displacement?

24. A particle starts with S.H.M. from the mean position as shown in the figure. Its amplitude is \(A\) and its time period is \(T\). At one time, its speed is half that of the maximum speed. What is its displacement?

(a) \( \frac{A}{2} \)

(b) \( \frac{A}{\sqrt{2}} \)

(c) \( \frac{A}{4} \)

(d) \( \frac{A\sqrt{3}}{2} \)

Detailed Answer:

In Simple Harmonic Motion (SHM), the relationship between speed \(v\), displacement \(x\), and maximum speed \(v_{\text{max}}\) is given by: \[ v = v_{\text{max}} \sqrt{1 - \left( \frac{x}{A} \right)^2} \] where: - \(v_{\text{max}} = A\omega\) is the maximum speed, - \(A\) is the amplitude of oscillation, - \(x\) is the displacement, - \(\omega\) is the angular frequency. We are told that the speed at a certain point is half of the maximum speed: \[ v = \frac{v_{\text{max}}}{2}. \] Substituting into the equation for speed: \[ \frac{v_{\text{max}}}{2} = v_{\text{max}} \sqrt{1 - \left( \frac{x}{A} \right)^2}. \] Dividing both sides by \(v_{\text{max}}\): \[ \frac{1}{2} = \sqrt{1 - \left( \frac{x}{A} \right)^2}. \] Squaring both sides: \[ \frac{1}{4} = 1 - \left( \frac{x}{A} \right)^2. \] Rearranging: \[ \left( \frac{x}{A} \right)^2 = \frac{3}{4}. \] Taking the square root: \[ \frac{x}{A} = \frac{\sqrt{3}}{2}. \] Therefore, the displacement \(x\) is: \[ x = A \cdot \frac{\sqrt{3}}{2}. \] The correct option is (d)

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