A simple harmonic motion of amplitude A has time period T. what will the velocity of the oscillator at (i) equilibrium position.
(ii) half amplitude

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Solution

$A m p l i t u d e = A t i m e p e r i o d = T a n g u l a r f r e q u e n c y = ω = \frac{2 π}{T} d i s p l a c e m e n t i n t h e S H M i s g i v e n b y : y = A \sin ω t v e l o c i t y i s g i v e n b y : v = \frac{d y}{d t} = A ω \cos ω t = ω \sqrt{A^{2} - y^{2}} a t e q u i l i b r i u m p o i s t i o n, y = 0 v e l o c i t y i s g i v e n b y = A ω = \frac{2 A π}{T} a t h a l f o f t h e a m p l i t u d e, y = A / 2 v = ω \sqrt{A^{2} - A^{2} / 4} v = \frac{\sqrt{3} A π}{4 T}$

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A simple harmonic motion of amplitude A has time period T. what will the velocity of the oscillator at (i) equilibrium position.(ii) half amplitude

Amplitude = Atime period = Tangular frequency = ω = 2πTdisplacement in the SHM is given by:y = A sinωtvelocity is given by:v= dydt= Aωcosωt = ωA2-y2at equilibrium poistion, y = 0velocity is given by = Aω = 2AπTat half of the amplitude, y =A/2v = ωA2-A2/4v =3Aπ4T

A simple harmonic motion of amplitude A has time period T. what will the velocity of the oscillator at (i) equilibrium position.
(ii) half amplitude