A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ${\mathrm{ω}}_0\text{. }$. An external force F(f) proportional to cos $\mathrm{ωt}\left(\mathrm{ω}≠{\mathrm{ω}}_0\right)$is applied to the oscillator. The time displacement of the oscillator will be proportional to

detailed solution

Correct option is B

Given that natural angular frequency =ω0 Let at any instant t, angular frequency =ω At a displacement x, the resultant accelerationf=ω02−ω2x External forec. F=mω02−ω2x Given that F∝cos⁡(ωt) âˆ´ mω02−ω2x∝cos⁡(ωt)  We know that x=Asin⁡(ωt+ϕ) Where the symbols have their usual meanings' Now A=Asin⁡(0+ϕ) (∵ at t=0,x=A) âˆ´ Ï•=π/2 thererefore x=Asin⁡ωt+π2=Acos⁡ωt From eqs. (3) and [5), we gst mω02−ω2Acos⁡ωt∝cos⁡ωt or A∝1mω02−ω2