Question

If the minimum time taken by a particle executing simple harmonic motion to move from extreme position to a point at a displacement of 86.6% of the amplitude is T, then the minimum time taken by the particle to move from mean position to a point at a displacement of 86.6% of the amplitude is:

• A. T
• B. 0.5 T
• C. 0.25 T
• D. 2 T

Hint:

A standard reference circle (phasor diagram) makes this effortless. Moving from the extreme position to \( \frac{\sqrt{3}}{2}A \) sweeps out a phase angle of \( 30^\circ \) (\( \omega T = \pi/6 \)). Moving from the mean position to \( \frac{\sqrt{3}}{2}A \) sweeps out an angle of \( 60^\circ \) (\( \omega t' = \pi/3 \)). Since the phase angle doubles, the time taken must also double!

Answer 1

The Correct Option is D

Concept: The displacement equation for a particle executing simple harmonic motion (S.H.M.) depends on its reference starting point:

• When measured from the **mean position**, the displacement is given by: \( x(t) = A\sin(\omega t) \)

• When measured from the **extreme position**, the displacement is given by: \( x(t) = A\cos(\omega t) \)
Here, \( A \) is the amplitude, \( \omega \) is the angular frequency, and the given displacement value 86.6% corresponds to the exact fraction \( \frac{\sqrt{3}}{2}A \) since \( \frac{\sqrt{3}}{2} \approx 0.866 \).

Step 1: Analyzing motion starting from the extreme position.
The particle moves from the extreme position (\( x = A \)) to a point where the displacement is \( 86.6\% \) of the amplitude, meaning \( x = \frac{\sqrt{3}}{2}A \), in a minimum time interval \( T \): \[ x(t) = A\cos(\omega t) \implies \frac{\sqrt{3}}{2}A = A\cos(\omega T) \] Cancelling \( A \) from both sides: \[ \cos(\omega T) = \frac{\sqrt{3}}{2} \implies \omega T = \frac{\pi}{6} \implies T = \frac{\pi}{6\omega} \quad \cdots (1) \]

Step 2: Analyzing motion starting from the mean position.
Let \( t' \) be the minimum time taken by the particle to travel from the mean position (\( x = 0 \)) to the same displacement point \( x = \frac{\sqrt{3}}{2}A \): \[ x(t) = A\sin(\omega t) \implies \frac{\sqrt{3}}{2}A = A\sin(\omega t') \] Cancelling \( A \) from both sides: \[ \sin(\omega t') = \frac{\sqrt{3}}{2} \implies \omega t' = \frac{\pi}{3} \implies t' = \frac{\pi}{3\omega} \quad \cdots (2) \]

Step 3: Relating \( t' \) to \( T \).
Dividing equation (2) by equation (1) to express the target time in terms of \( T \): \[ \frac{t'}{T} = \frac{\frac{\pi}{3\omega}}{\frac{\pi}{6\omega}} = \frac{6}{3} = 2 \implies t' = 2T \] Thus, the minimum time taken to move from the mean position is exactly \( 2T \).