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Question

Consider a particle moving in simple harmonic motion according to the equation x=2.0cos(50Π t+tan10.75)

where x is in centimeter and t in second. The motion is started at t = 0. (a) When does the particle come to rest for the second time?



Correct Answer
A

sec

Your Answer
B

sec

Your Answer
C

sec

Your Answer
D

sec


Solution

The correct option is A

sec


x=2.0cos(50Π t+tan10.75)

v=dxdt=π sin(50Π t+tan10.75) 

  Find t when v=0

=100π sin(50π t+tan10.75)

sinθ=0 

θ=nπ

θ=((50π t+tan10.75)=nπ

t=nπtan10.7550π    (tan10.75=0.64)

First time particle will come to rest after start (i.e., t=0) is when π=1. second time it will happen when n

=2

t=2π0.6450π   ( tan10.75=0.64)

n=2  t=2π0.6450π

=3.6×102sec

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