First slide

Simple harmonic motion

Question

Time period (T) and amplitude (A) are same for two particles which undergo SHM along the same line. At one particular instant, one particle is at phase $\frac{3 π}{2}$ and other is at phase zero. While moving in the same direction. Find the time at which they will cross each other:

Moderate

A

4T/2
B

3T/8
C

3T/4
D

3T/7

Solution

$At phase 3 π / 2, (ω t + ϕ) = 3 π / 2$
$\begin{array}{l} x_{1} = A \sin \frac{3 π}{2} = - A \\ x_{2} = 0 \end{array}$
$So, at t = 0, x_{1} = - A \cos ω t and x_{2} = A \sin ω t$
$\begin{array}{l} x_{1} = x_{2} \\ - \cos ω t = \sin ω t \Rightarrow \tan ω t = - 1 \\ ω t = \frac{3 π}{4} \Rightarrow (\frac{2 π}{T}) t = \frac{3 π}{4} or t = 3 T / 8 \end{array}$

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