First slide
Simple hormonic motion
Question

A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω0. An external force F(f) proportional to cos ωtωω0is applied to the oscillator. The time displacement of the oscillator will be proportional to

Difficult
Solution

Given that natural angular frequency =ω0 Let at any instant t, angular frequency =ω At a displacement x, the resultant acceleration
f=ω02ω2x External forec. F=mω02ω2x Given that Fcos(ωt)  mω02ω2xcos(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)  ϕ=π/2 thererefore x=Asinωt+π2=Acosωt From eqs. (3) and [5), we gst mω02ω2Acosωtcosωt or A1mω02ω2

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