Calculating Qf: Piano Key Frequency 256Hz Energy Dissipated in 1s

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SUMMARY

The discussion focuses on calculating the effective quality factor (Qf) of a piano key oscillating at a frequency of 256 Hz, which dissipates energy to half its initial value in 1 second. The relevant equations used are Qf = ω/v and = E_{0} e^(-vt). The calculated value of v is 0.693, leading to an effective Qf of 369. The solution confirms the correct application of the energy dissipation formula and the conversion to angular frequency.

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Homework Statement


"Systems typically exhibit an exponential decrease in their average stored energy of the form <E> = E_{0} e^(-vt) ---- If a piano key of frequency 256 Hz is struck and its oscillation energy decreases to one half of its initial value in about 1 second what is the effective Qf of the system?


Homework Equations



Qf = w/v
<E> = E_{0} e^(-vt)


The Attempt at a Solution



I need to find the energy dissipated and from the equation they gave me I put 1/2 (the energy the system has from its initial value after 1 second)

1/2 = e^(-v)

v = .693

Qf = 256/.693 = 369

I'm not quite sure if I did the dissipation energy equation correctly.
 
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Q=\frac{\omega}{v}=\frac{2 \pi f}{v}. Your numerator needs to be in angular frequency. Looks fine though.
 

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