Physics, asked by ankitjena75, 8 months ago

A particle executes simple harmonic motion and when its displacement from the mean position is 2.5 cm, its kinetic and potential energies are of equal magnitude. The amplitude of the motion of the particle in cm is approximately: (a) 2.5

(b) 3.5

(c) 5.0

(d) 7.5

Answers

Answered by deepthimuthu77
1

Answer:

A particle oscillates with simple harmonic motion, so that its displacement varies according to the expression x = (5 cm)cos(2t + π/6) where x is in centimeters and t is in seconds. At t = 0 find (a) the displacement of the particle, (b) its velocity, and (c) its acceleration. (d) Find the period and amplitude of the motion.

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