Question

A damped osillator has a mass (m = 2.00kg), a spring (k = 10.0N/m), and a damping coefficient b = 0.102kg/s. undamped angular frequency of the system is 2.24rad/s. If the initial amplitude is 0.250m, How many periods of motion are necessary for the amplitude to be reduced to 3/4 it initial value? is this system underdamped, critically damped, or overdamped

Answer 1

mass=m=2 kg

spring constant=k=10 N/m

b=0.102 kg/s

undamped angular frequency=w0=2.24 rad/s

initial amplitude=0.25 m

the governing equation for the oscillator can be written as:

m*d^2x/dt^2 + c*dx/dt + k*x=0

==> 2*d^2x/dt^2 + 0.102*dx/dt + 10*x=0

let x=exp(A*t)

then 2*A^2+0.102*A+10=0

==>A=-0.0255+2.236 i

and A=-0.0255-2.236 i

as the roots are imaginary, the system is underdamped in nature.

solution for x(t) can be written as

x(t)=exp(-0.0255*t)*(C*cos(2.236*t)+D*sin(2.236*t))

where C and D are two constants.

given that at t=0, x(t)=0.25 m

C=0.25 m

dx/dt=exp(-0.0255*t)*(-2.236*C*sin(2.236*t)+2.236*D*cos(2.236*t))+(C*cos(2.236*t)+D*sin(2.236*t))*(-0.0255*exp(-0.0255*t))

at t=0, speed=0

==>dx/dt=0

==>2.236*D+C*(-0.0255)=0

==>D=2.85*10^(-3)

hence complete solution is given as

x(t)=exp(-0.0255*t)*(0.25*cos(2.236*t)+2.85*10^(-3)*sin(2.236*t))

as seen from the equation, the amplitude gets reduced by exp(-0.0255*t)

so to get the amplitude reduced to 3/4 times,

3/4=exp(-0.0255*t)

==>t=11.28 seconds

time period=2*pi/(2.24)=2.805 seconds

number of oscillations=11.28/2.805=4 oscillations