Q7.18 Damping (T)

The figure shows the displacement (x, in meters) as a function of time (t, in seconds) for an oscillating object.

Describe the behaviour in words.

Estimate:

  1. The SHM frequency
  2. The time for the amplitude to decay to 1% of its initial value.
  3. The time for the total energy to decay to 1% of its initial value.
Graphic - No title - ampdampednums
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Hint

How does the total energy depend on the amplitude of the motion?
 

Solution

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Solution

The figure shows damped simple harmonic motion. One can think of the system oscillating in a conventional SHM fashion, but with an amplitude that is slowly decaying with time.
  1. The time taken for one cycle can be estimated from the figure as T=1s so the SHM frequency is 1 Hz.
  2. The figure suggests (what we saw in S6.6) that the amplitude decays exponentially with time:
    xm(t)=xm(0)e-λt
    where λ depends on the strength of the damping. An eyeball estimate suggests that the amplitude decays from xm=1 to xm=0.25 in time t=7 s. Thus
    \[ -t \lambda= \ln \left[ \frac{x_m(t)}{x_m(0)} \right] \hspace{0.25cm}\mbox{\rm implying}\hspace{0.25cm} \lambda = - \ln [0.25]/7 = 0.2 \]
    The time for the amplitude to decay to 1% of its initial value is the solution to
    \[ 0.01 = e^{-\lambda t} \hspace{0.25cm}\mbox{\rm or}\hspace{0.25cm} t = - \frac{\ln(0.01)}{\lambda} = \frac{4.6}{0.2}= 23\,s \]
  3. The total energy associated with SHM is proportional to the square of the amplitude. Thus

    \[ \frac{E(t)}{E(0)} = \frac{x_m^2(t)}{x_m^2(0)} = \left [ e^{-\lambda t} \right] ^2 = e^{-2\lambda t} \]

    [If you got that last step wrong then review the properties of exponentials!]

    The time for the energy to decay to 1% of its initial value is thus the solution to

    \[ 0.01 = e^{-2\lambda t} \hspace{0.25cm}\mbox{\rm giving}\hspace{0.25cm} t=\frac{4.6}{0.4}= 11.5\,s \]