Two particles are in SHM with same amplitude A and same angular frequency ω . At time t = 0 , one is at x = + 2 A and other is at x = − 2 A . Both are moving in same direction.

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PhysicsSHMMiscellaneousMedium2 minQB

PhysicsMediummultiple choice

Two particles are in SHM with same amplitude A and same angular frequency $ω$ . At time $t = 0$ , one is at $x = + \frac{A}{2}$ and other is at $x = - \frac{A}{2}$ . Both are moving in same direction.

Options:(select one or more)

Answer:

A, C

Solution:

Both particles start at $x = \pm A /2$ moving in the same direction, so their phase angles differ by $Δ ϕ = \frac{π}{3}$ (A correct, B wrong).

They meet when the faster phasor gains 180° relative geometry, i.e. the phasor must rotate by $π /2$ from the initial configuration: $t = \frac{π /2}{ω} = \frac{π}{2 ω}$ so (C) is correct and (D) is wrong.

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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