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

Step 1: Set up the two SHM equations.
When we count from the mean position, displacement is $x = A\sin\omega t$. When we count from the extreme position, displacement is $x = A\cos\omega t$. We will use both.
Step 2: Convert the percentage.
The displacement $86.6\%$ of amplitude means $x = 0.866A$, which is exactly $\dfrac{\sqrt{3}}{2}A$ because $\dfrac{\sqrt{3}}{2} \approx 0.866$.
Step 3: Handle the motion from the extreme position.
Starting from the extreme, reaching $x = \dfrac{\sqrt{3}}{2}A$ takes time $T$: \[ \frac{\sqrt{3}}{2}A = A\cos(\omega T) \;\Rightarrow\; \cos(\omega T) = \frac{\sqrt{3}}{2} \] So $\omega T = \dfrac{\pi}{6}$, giving $T = \dfrac{\pi}{6\omega}$.
Step 4: Handle the motion from the mean position.
Let the time be $t'$. Starting from the mean, reaching the same point: \[ \frac{\sqrt{3}}{2}A = A\sin(\omega t') \;\Rightarrow\; \sin(\omega t') = \frac{\sqrt{3}}{2} \] So $\omega t' = \dfrac{\pi}{3}$, giving $t' = \dfrac{\pi}{3\omega}$.
Step 5: Divide one time by the other.
\[ \frac{t'}{T} = \frac{\pi/(3\omega)}{\pi/(6\omega)} = \frac{6}{3} = 2 \]
Step 6: State the result.
So the time from the mean position is twice the time from the extreme: \[ \boxed{t' = 2T} \]