For a damped simple harmonic oscillator, the block has a mass of 2.2 kg and the

Question

For a damped simple harmonic oscillator, the block has a mass of 2.2 kg and the spring constant is 6.0 N/m. The damping force is given by -b(dx/dt), where b = 160 g/s. The block is pulled down 11.9 cm and released. (a) Calculate the time required for the amplitude of the resulting oscillations to fall to 1/5 of its initial value. (b) How many oscillations are made by the block in this time?

For a damped simple harmonic oscillator, the block has a mass of 2.2 kg and the spring constant is 6.0 N/m. The damping force is given by -b(dx/dt), where b = 160 g/s. The block is pulled down 11.9 cm and released. (a) Calculate the time required for the amplitude of the resulting oscillations to fall to 1/5 of its initial value. (b) How many oscillations are made by the block in this time?

Explanation / Answer

a) Net force is given as

fnet = -kx - b (dx/dt)

Now x(t)=xm/5 and x(t) is given as

x(t) = xm * exp(-bt/2m)

xm/5= xm * exp(-bt/2m)

>>>>>>>bt/2m = ln5

t = ln(5)*2*2200/ (160) = 44.26 sec

b) The angular velocity is given as

? = sqrt ( k/m + b^2/4m^2 )

=sqrt ( 6000/2200 + 160^2/4*2200^2 )

=1.65 rad/sec

Number of oscillations =t/T = t?/2? = 44.26*1.65/2?

n=11.63

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