Question

A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is 5 sec . The time required by it to move from $x=A$ to $x=\frac{A}{\sqrt{2}}$ is ________ sec.

• A. 1/4
• B. 5/4
• C. 5/8
• D. 3/8

Hint:

Write $x = A\cos\omega t$ and find the time at which $x = A/\sqrt{2}$; compare it with the time period $T$.

Answer 1

The Correct Option is C

Use the reference-circle (phasor) picture: the SHM is the projection of uniform circular motion of radius $A$ whose period is also $T$.
Step 1: The particle starts at $x=A$, i.e. at angle $\theta = 0$. The position $x = A\cos\theta$ equals $\frac{A}{\sqrt{2}}$ when $\theta = 45^\circ$.
Step 2: The phasor turns through $360^\circ$ in a time $T$, so it turns through $45^\circ$ in the fraction $\frac{45^\circ}{360^\circ} = \frac{1}{8}$ of the period.
Step 3: $t = \frac{T}{8} = \frac{5}{8}$ s.
The required time is $5/8$ s, which is option (3).