Question

A damped oscillator is formed by attaching a mass with m = 1.5 kg to one end of a spring with spring constant k = 8 N/m. The other end of the spring is anchored and the mass can slide on a horizontal surface The damping force is given by –bv with b = 230 g/s. At t=0, the mass is displaced so that the spring is compressed by 12 cm from its unstretched length and released from rest.

(a) Find the time required for the amplitude of the resulting oscillations to decay to 1/3 of its initial value.

(b) How many oscillations are made by the mass during this time?

(c) Find the value of b so that the oscillator is critically damped.

(d) At t=0, this critically damped oscillator is displaced so that the spring is stretched a distance of 12 cm beyond its unstretched length, find the time required for mass to reach the position for which the spring is stretched by only 4 cm.

Answer 1

(a) maximum Amplitude x(t) with damping is given by   x(t) = x_m e^-bt/(2m)

If x(t) = (1/3) x_m = x_m e^-bt/(2m) , then bt/2m = ln(3) , hence t = ln(3) (2m)/b

t = 1.099 2 1.5 / ( 230 10^-3 ) = 14.3 s

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angular velocity with damping is given by

period of oscillation T is given by

Number of oscillations before amplitude becoming one-third of initial amplitude = 14.3 / 2.72 5

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for critical damping , b = 2(km)^1/2 = 2 ( 8 1.5 )^1/2 = 6.93 kg/s

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x = x_me^-bt/(2m)

4 = 12 e^-bt/(2m)   or e^-bt/(2m) = 1/3

Hence t = ln(3)(2m)/b = 1.099 2 1.5 / 6.93 = 0.475 s

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