First slide

Simple harmonic motion

Question

Two particles P and Q describe SHM of same amplitude a, same frequency f along the same straight line. The maximum distance between the two particles is $a \sqrt{2}$ .The phase difference between the particle is

Moderate

A

zero
B

$\frac{π}{2}$
C

$\frac{π}{6}$
D

$\frac{π}{3}$

Solution

$\begin{array}{l} x_{1} = a \sin (ω t + ϕ_{1}) \\ x_{2} = a \sin (ω t + ϕ_{2}) \\ \Rightarrow |x_{1} - x_{2}| = 2 a \sin (ω t + \frac{ϕ_{1} + ϕ_{2}}{2}) \cos (\frac{ϕ_{1} - ϕ_{2}}{2}) \\ To maximize |x_{1} - x_{2}| : \sin (ω t + \frac{ϕ_{1} + ϕ_{2}}{2}) = 1 \\ \Rightarrow a \sqrt{2} = 2 a \times 1 \times \cos (\frac{ϕ_{1} - ϕ_{2}}{2}) \\ \Rightarrow \frac{1}{\sqrt{2}} = \cos (\frac{ϕ_{1} - ϕ_{2}}{2}) \end{array}$
$\Rightarrow \frac{π}{4} = \frac{ϕ_{1} - ϕ_{2}}{2} \Rightarrow ϕ_{1} - ϕ_{2} = \frac{π}{2}$

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