Two particles are performing simple harmonic motion in a straight...

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to

A

and

T

, respectively. At time

t = 0

one particle has displacement

A

while the other one has displacement

- \dfrac{A}{2}

and they are moving towards each other. If they cross each other at time

t

, then

t

is:

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude

Options

A $\dfrac{5T}{6}$
B $\dfrac{T}{3}$
C $\dfrac{T}{4}$
D $\dfrac{T}{6}$

Correct answer

D. $\dfrac{T}{6}$

Step-by-step solution

Angle covered to meet

= {60}^{o} = \dfrac{\pi}{3}rad

Therefore,

t = \dfrac{\theta}{\omega} = \dfrac{\pi}{3} \times \dfrac{T}{2\pi} = \dfrac{T}{6}

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