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Question

A simple harmonic oscillator of angular frequency 2 rad/s is acted upon by an external force F=sint N. If the oscillator is at rest in its equilibrium position at t=0, its position at later times is proportional to:


  1. sint+12cos2t
  2. cost12sin2t
  3. sint+12sin2t
  4. sint12sin2t

A
sint+12cos2t
B
sint+12sin2t
C
sint12sin2t
D
cost12sin2t
Solution
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The correct option is D sint12sin2t
According to the question,
Given; F=Fsinta=Fmsintv=Fmcost
x=FmsintX=rx1+rx2=Asin2tFmsint..............................eqi

Now, we know that in SHM,
a (acceleration)=ω2Asinωt and above we find a=Fmsint

Resultant force, Fresultant=sintmω2Asinwt

FR=sint4mAsin2t (Given, ω=2radian/s)
aR=sint/m4Asin2t (By integrating)
vR=cost/m+2Acos2t (By integrating)
xR=sint/m+Asin2t

We can write, xR=1m(sintAmsin2t) xR(sintAmsin2t)

Only option D is in the given form, Hence, option D is correct.

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