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:
sint+12cos2t
cost−12sin2t
sint+12sin2t
sint−12sin2t
A
sint+12cos2t
B
sint+12sin2t
C
sint−12sin2t
D
cost−12sin2t
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Solution
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The correct option is Dsint−12sin2t According to the question,
a(acceleration)=−ω2Asinωtand above we find a=Fmsint
Resultant force, Fresultant=sint−mω2Asinwt
⇒FR=sint−4mAsin2t (Given, ω=2radian/s)
⇒aR=sint/m−4Asin2t (By integrating)
⇒vR=−cost/m+2Acos2t(By integrating)
⇒xR=−sint/m+Asin2t
We can write, xR=−1m(sint−Amsin2t)⇒xR∝(sint−Amsin2t)
Only option D is in the given form, Hence, option D is correct.
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