First slide

Simple harmonic motion

Question

Two particles move parallel to the x-axis about the origin with same amplitude 'a' and frequency $ω$ . At a certain instant they are found at a distance a/3 from the origin on opposite sides but their velocities are in the same direction. What is the phase difference between the two?

Moderate

A

$\cos^{- 1} \frac{7}{9}$
B

$\cos^{- 1} \frac{5}{9}$
C

$\cos^{- 1} \frac{4}{9}$
D

$c {os}^{- 1} \frac{1}{9}$

Solution

$Let x_{1} = a \sin ω t and x_{2} = a \sin (ω t + δ) be two SHMs.$
$\begin{array}{l} \frac{a}{3} = a \sin ω t and - \frac{a}{3} = a \sin (ω t + δ) \\ \sin ω t = 1 / 3 and \sin (ω t + δ) = - 1 / 3 \end{array}$
$Eliminating t, \frac{1}{3} \cos δ + \sqrt{1 - \frac{1}{9}} \sin δ = - \frac{1}{3}$
$\begin{array}{l} 9 \cos^{2} δ + 2 \cos δ - 7 = 0 \\ \cos δ = - 1 or \frac{7}{9} \\ i . e ., δ = 180^{\circ} or \cos^{- 1} (\frac{7}{9}) \end{array}$
If we put 180^o, we find that v₁ and v₂ are of opposite signs.
Hence $δ$ = 180^o is not applicable.
$∴ δ = \cos^{- 1} (\frac{7}{9})$

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