Lightly Damped Simple Harmonic Oscillator

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SUMMARY

The discussion focuses on the analysis of a lightly damped simple harmonic oscillator (SHO) using a tuning fork with a natural frequency of 392 Hz. The angular frequency is calculated as 2463 rad/s. The damping factor (gamma) is determined to be 0.23, derived from the amplitude decay over 10 seconds, where the amplitude reduces to 10% of its original value. The formula used for amplitude as a function of time is A(t) = A₀ e^{-tγ}, confirming the calculations through logarithmic manipulation.

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DrDank
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Tuning forks are lightly damped SHO's. Consider a tuning fork who's natural frequency is f=392Hz. Angular frequency = w = 2(Pi)f = 2463 (rad/s)

The damping of this tuning fork is such that, after 10 sec, it's amplitude is 10% of it's original amplitude.

Here is my attempt to find the damping factor (gamma)
Amplitude as a function of time where g is the damping factor (g = gamma)A(t) = A_{o} e^{-t\gamma}

A(10) = \frac{1}{10} A(0)

A_{o}e^{-10\gamma} = \frac{1}{10} A_{o} e^{0}

e^{-10\gamma} = \frac{1}{10}

-10 \gamma = \ln{\frac{1}{10}}

\gamma = \frac{\ln{10}}{10} = .23

Is this right?
 
Last edited:
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You can check your result yourself by substituting it back into the original equation. :smile:
 

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