Let F=F0sinwt is the external periodic force acting on the oscillator. If amplitude of the oscillator is maximum for w=w1 and the energy is maximum for w=w2,then (w0 is the natural angular frequency)the relation b/w w1,w2 and w0 is)

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Solution

$D e a r s t u d e n t F = F_{0} \sin ω t n a t u r a l a n g u l a r f r e q u e n c y = ω_{0} s o, F = F_{0} \sin ω_{0} t ω_{0} = \frac{2 π}{T} - - - - - - - - 1 m a x a m p l i t u d e = F_{0} F = F_{0} w h e n \sin ω_{1} t = 1 o r ω_{1} t = π / 2 o r, t = \frac{π}{2 ω_{1}} o r, t = \frac{T}{4} n o w, 1 = \sin ω_{1} \frac{T}{4} \sin \frac{π}{2} = \sin ω_{1} \frac{T}{4} \frac{π}{2} = ω_{1} \frac{T}{4} \frac{2 π}{T} = ω_{1 - - - - - - 2} K E = \frac{1}{2} m ω^{2} {F_{0}}^{2} \cos^{2} ω t P E = \frac{1}{2} m ω^{2} {F_{0}}^{2} \cos^{2} ω t E_{t o t a l} = \frac{1}{2} m ω^{2} {F_{0}}^{2} \cos^{2} ω t + \frac{1}{2} m ω^{2} {F_{0}}^{2} \cos^{2} ω t = \frac{1}{2} m ω^{2} {F_{0}}^{2} E_{t o t a l} i s m a x w h e n F_{0} h a s h i g h e s t v a l u e h e r e, ω = ω_{1} = ω_{2} a s p e r q u e s t i o n s o, ω_{0} = ω_{1} = ω_{2}$

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