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Question: A particle of mass m is attached to a spring (of spring constant k) and has a natural angular freque...

A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω0{{\omega }_{0}}. An external force F(t) proportional to cosωt(ωω0)\cos \omega t\left( \omega \ne {{\omega }_{0}} \right)is applied to the oscillator. The maximum displacement of the oscillator will be proportional to
A.mω02ω2\dfrac{m}{{{\omega }_{0}}^{2}-{{\omega }^{2}}}
B.1m(ω02ω2)\dfrac{1}{m\left( {{\omega }_{0}}^{2}-{{\omega }^{2}} \right)}
C.1m(ω02+ω2)\dfrac{1}{m\left( {{\omega }_{0}}^{2}+{{\omega }^{2}} \right)}
D.mω02+ω2\dfrac{m}{{{\omega }_{0}}^{2}+{{\omega }^{2}}}

Explanation

Solution

As a first step, you could read the question carefully and hence note down every point that seems important from it. Then you could recall the expression for displacement for an SHM. Then you could work accordingly to find the proportionality relation for maximum displacement of this oscillator.

Formula used:
Displacement of an SHM,
x=Asin(ωt+ϕ)x=A\sin \left( \omega t+\phi \right)

Complete step-by-step solution:
Let the angular velocity at any instant be given by ω\omega and the initial angular velocity of the particle also given to be ω0{{\omega }_{0}}. The resultant acceleration for a displacement x could be given by,
a=(ω02ω2)xa=\left( {{\omega }_{0}}^{2}-{{\omega }^{2}} \right)x
Now, the external force corresponding to this acceleration could be given by,
F=m(ω02ω2)xF=m\left( {{\omega }_{0}}^{2}-{{\omega }^{2}} \right)x…………………………………. (1)
We are given in the question that,
FcosωtF\propto \cos \omega t
From (1) we have,
m(ω02ω2)xcosωtm\left( {{\omega }_{0}}^{2}-{{\omega }^{2}} \right)x\propto \cos \omega t………………………………. (2)
Now, the expression for displacement of a simple harmonic motion could be given by,
x=Asin(ωt+ϕ)x=A\sin \left( \omega t+\phi \right)
Now, for time t=0 and x=A, we have,
A=Asin(0+ϕ)A=A\sin \left( 0+\phi \right)
ϕ=π2\Rightarrow \phi =\dfrac{\pi }{2}
x=Asin(ωt+π2)=Acosωt\Rightarrow x=A\sin \left( \omega t+\dfrac{\pi }{2} \right)=A\cos \omega t
Substituting this in (2) we would get,
m(ω02ω2)Acosωtcosωtm\left( {{\omega }_{0}}^{2}-{{\omega }^{2}} \right)A\cos \omega t\propto \cos \omega t
A1m(ω02ω2)\therefore A\propto \dfrac{1}{m\left( {{\omega }_{0}}^{2}-{{\omega }^{2}} \right)}
Therefore, the maximum displacement of the oscillator will be proportional to 1m(ω02ω2)\dfrac{1}{m\left( {{\omega }_{0}}^{2}-{{\omega }^{2}} \right)}. Hence, option B is correct.

Note: There are many small definitions that come under the concept of waves that should be known by heart by the students for a better understanding of higher concepts. Here, one should know that maximum displacement of an oscillator stands for the amplitude of the wave otherwise it is impossible to answer this question.