Two particles in SHM share a mean position, amplitude A and period T . At t = 0 one is at + A and the other is at 2 3 A on the negative side, moving towards the mean position. Find the time when they first cross, as a multiple of T .

x 2 = A cos ( ω t − 5 π /6 ) . Equal when ω t = 5 π /12 , t = 24 5 T .

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PhysicsSHMOscillations and SHMMedium2 min

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Two particles in SHM share a mean position, amplitude $A$ and period $T$ . At $t = 0$ one is at $+ A$ and the other is at $\frac{3}{2} A$ on the negative side, moving towards the mean position. Find the time when they first cross, as a multiple of $T$ .

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Answer:

Solution:

$x_{2} = A cos (ω t - 5 π /6)$ . Equal when $ω t = 5 π /12$ , $t = \frac{5 T}{24}$ .

Subject:PhysicsTopic:SHMSubtopic:Oscillations and SHM

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