Question

A particle undergoes simple harmonic motion having time period T. Find the time taken in 3/8th oscillation, if the particle starts from mean position.

Answer 1

The speed of a particle executing SHM isn’t constant, but it decreases as the particle moves from mean position to extreme and increases when the particle moves from extreme to mean position. In a complete oscillation, a particle travels a distance 4A, hence 3/8th oscillation would mean a distance of 3A/2 distance, where A denotes the amplitude of the particle in SHM.

Complete step by step solution:
Since the particle is starting from mean position and it needs to cover a distance of 3A/2, we can divide the distance into two parts

Part 1 : A distance A is travelled when particle moves from mean position to extreme position. During this motion the particle will take time T/4.

Part 2 : A distance A/2 is travelled from extreme position towards mean position. Now we need to calculate the time required to cover this journey.
Considering the equation of the particle starting from mean position to be
$y = A\operatorname{Sin} \omega t$,
Putting : $t = \dfrac{T}{4}$ , and $\omega = \dfrac{{2\pi }}{T}$ , we can prove that the particle reaches the positive extreme in time T/4.
Now the particle will reach the point $y = \dfrac{A}{2}$ , at time calculated form the equation
$\dfrac{A}{2} = A\operatorname{Sin} \omega t$,
$\dfrac{1}{2} = \operatorname{Sin} \left( {\dfrac{{2\pi }}{T}t} \right)$,
Looking for solution between $t = \dfrac{T}{4}$, and $t = \dfrac{T}{2}$, we get
$\dfrac{{2\pi }}{T}t = \dfrac{{5\pi }}{6}$,
$t = \dfrac{{5T}}{{12}}$.
Hence starting from a mean position at $t = 0$, the particle covers 3A/2 distance at time $t = \dfrac{{5T}}{{12}}$.
This means that the particle takes $\dfrac{{5T}}{{12}}$ time to complete 3/8th oscillation.

Note: To solve such problems, a student must use correct form of the equation, i.e. if a particle starts from mean position, we can use the form $y = A\operatorname{Sin} \omega t$, but if the particle starts from extreme position, we will use the form $y = A\operatorname{Cos} \omega t$.