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Grade: 12th pass
        A simple harminic oscillator has a period of 0.01sec.and an amplitude of 0.2m.the speed in m/s.at the center of oscillation is
10 months ago

Answers : (1)

Arun
14736 Points
							
Dear student
 
Start by using this equation for the displacement x of the oscillator: 
x = A*sin(w*t) where A is the amplitude, w is the angular frequency, and t is time. 
speed = dx/dt = A*w*cos(w*t) 
From your data A = 0.2m, & w = 2*pi/T = 628rad/s (T is the period). The speed at the centre of the oscillation is the speed when t = 0, i.e. when cos(w*t) = 1 
speed at centre = 0.2*628 = 125.7m/s = 40*pi m/s

Regards
Arun (askIITians forum expert)
10 months ago
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