Question

Damped SHM motion A mass M is suspended from a spring and oscillates with a period of 0.840 s. Each complete oscillation results in an amplitude reduction of a factor of 0.965 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 69% of its initial value. The amplitude after N oscillations- (initial amplitude) x(damping factor)N Submit wIncorrect. Tries 2/6 Previous Tries 1998-2018 by Florida State University. All rights reserved The contents of this web site are for use by the students enrolled in this course only. Any distribution of the contents in whole or in parts and in any form, including posting on the web or in printed version, is a violation of the Copyright laws and the FSU honor codes.

Answer 1

Let the initial amplitude of the SHM be A₀ . The energy of SHM is proportional to square of the amplitude. Therefore the initial energy E₀ is,

k is the constant of proportionality,

After N oscillations the amplitude wil be,

Therefore the Energy after N oscillations is,

Now, for E to become 0.69E₀ the number of oscillation required can be calculated by the equation,

$\Rightarrow 0.69E_0 = E_0(0.965)^{2N}$

$\Rightarrow 0.69 = (0.965)^{2N}$

$\Rightarrow log(0.69) = 2N log(0.965)$

$\Rightarrow N = 5.2$

The oscillation period is 0.840 s. This period remains same for all oscillations, so, the time taken is,

$\Rightarrow t = 0.840 \times 5.2 = 4.37 \, s$