Simple Harmonic Motion 3 Question 3

3. A particle executes simple harmonic motion with an amplitude of 5cm. When the particle is at 4cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time (in seconds) is

(2019 Main, 10 Jan II)

(a) 4π3

(b) 8π3

(c) 73π

(d) 38π

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Answer:

Correct Answer: 3. (b)

Solution:

  1. In simple harmonic motion, position (x), velocity (v) and acceleration (a) of the particle are given by

x=Asinωtv=ωA2x2 or v=Aωcosωt

 and a=ω2x or a=ω2Asinωt

Given, amplitude A=5cm and displacement x=4cm. At this time (when x=4cm ), velocity and acceleration have same magnitude.

|vx=4|=|ax=4| or |ω5242|=|4ω2|3ω=+4ω2ω=(3/4)rad/s

So, time period, T=2πωT=2π3×4=8π3s



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