Simple Harmonic Motion 1 Question 2

2. A particle undergoing simple harmonic motion has time dependent displacement given by x(t)=Asinπt90. The ratio of kinetic to potential energy of this particle at t=210s will be

(2019 Main, 11 Jan I)

(a) 2

(b) 1

(c) 19

(d) 3

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Answer:

Correct Answer: 2. (*)

Solution:

  1. Here given, displacement, x(t)=Asinπt90

where A is amplitude of S.H.M., t is time taken by particle to reach a point where its potential energy U=12kx2 and kinetic energy =12k(A2x2) here k is force constant and x is position of the particle.

Potential energy (U) at t=210s is

U=12kx2=12kA2sin221090π=12kA2sin22π+39π=12kA2sin2π3

Kinetic energy at t=210s, is

K=12k(A2x2)=12kA21sin2210π90=12kA2cos2(210π/90)

K=12kA2cos2(π/3)

So, ratio of kinetic energy to potential energy is

KU=12kA2cos2(π/3)12kA2sin2(π/3)=cot2(π/3)=13

No option given is correct.



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