Question

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

Answer 1

Hint: 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,
$x=A\sin \left( \omega t+\phi \right)$

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