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.

Question

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 $- \frac{A}{2}$ and they are moving towards each other. If they cross each other at time $t$ , then $t$ is:

Question diagram: Two particles are performing simple harmonic motion in a str

Options:

Answer:

D

Solution:

Angle covered to meet $= 60^{o} = \frac{π}{3} r a d$

Therefore, $t = \frac{θ}{ω} = \frac{π}{3} \times \frac{T}{2 π} = \frac{T}{6}$

TestHub · Accepted Answer

Angle covered to meet = 60 o = π 3 r a d Therefore, t = θ ω = π 3 × T 2 π = T 6

Physics - SHM Question with Solution | TestHub

Options:

Doubts & Discussion