Question

In terms of time period of oscillation $T$, what will be the shortest time in moving a particle from $+\frac{A}{2}$ to $-\frac{\sqrt{3}}{2}A$?

• A. $T$
• B. $\frac{T}{2}$
• C. $\frac{T}{4}$
• D. $\frac{3T}{4}$

Answer 1

The correct option is C $\frac{T}{4}$

As shown in figure, as SHM performing particle $P$ moves from $P_1$ to $P_2$, its corresponding point $Q$ moves on reference circle from $Q_1$ to $Q_2$.


In $△O{P}_1{Q}_1$
$\angle Q_{1}O{P}_1={\cos }^{-1}\frac{A/2}{A}$
$\Rightarrow \angle Q_{1}O{P}_1={\cos }^{-1}\frac{1}{2}$
$\Rightarrow \angle Q_{1}O{P}_1={60}^{\circ }$
Now in $△O{P}_2{Q}_2$
$\angle Q_{2}O{P}_2={\cos }^{-1}\frac{\sqrt{3}A/2}{A}$
$\Rightarrow \angle Q_{2}O{P}_2={\cos }^{-1}\frac{\sqrt{3}}{2}$
$\Rightarrow \angle Q_{2}O{P}_2={30}^{\circ }$
Angle between $Q_{1}O{Q}_2$ is
$\angle Q_{1}O{Q}_2={180}^{\circ }-({30}^{\circ }+{60}^{\circ })$
$\Rightarrow \angle Q_{1}O{Q}_2={90}^{\circ }$
The particle takes time $T$ to cover ${360}^{\circ }$
Time taken to complete ${90}^{\circ }=\frac{{90}^{\circ }T}{{360}^{\circ }}$
Time taken to complete ${90}^{\circ }=\frac{T}{4}$

Why this question? Tip:- Near the mean position, speed of the particle is more, therefore, it covers the first half of the distance (from $0$ to $A/2$) in a shorter time compared to the second half (from $A/2$ to $A$)