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 $$F\left( t \right)$$ proportional to $$\cos \,\omega t\left( {\omega \ne {\omega _0}} \right)$$ is applied to the oscillator. The displacement of the oscillator will be proportional to
A.
$$\frac{1}{{m\left( {\omega _0^2 + {\omega ^2}} \right)}}$$
B.
$$\frac{1}{{m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$
C.
$$\frac{m}{{\omega _0^2 - {\omega ^2}}}$$
D.
$$\frac{m}{{\left( {\omega _0^2 + {\omega ^2}} \right)}}$$
Answer :
$$\frac{1}{{m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$
Solution :
$$\eqalign{
& x = A\sin \left( {\omega t + \phi } \right) \cr
& {\text{where}}\,\,A = \frac{{{F_0}}}{{m\sqrt {{{\left( {\omega _0^2 - {\omega ^2}} \right)}^2}} }} = \frac{{{F_0}}}{{m\left( {\omega _0^2 - {\omega ^2}} \right)}} \cr} $$
Here damping effect is considered to be zero
$$\therefore x \propto \frac{1}{{m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$