Damped Harmonic Motion - Oscillating Spring

AI Thread Summary
A mountain bike with poor shock absorbers oscillates with a period of 0.5 seconds, and the mass of the bike and rider is 80 kg. The initial calculation for the spring constant k was found to be 12633 N/m, but this value is incorrect due to the quality factor (Q) being 10, which affects the natural frequency. The correct approach involves calculating the natural frequency using the relationship between k and m, as Q indicates that the system is not perfectly damped. The discussion emphasizes the need to adjust the calculations based on the quality factor to determine the accurate spring constant for the bike's shock absorbers. Accurate calculations are essential for understanding the bike's damping characteristics.
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Homework Statement



http://www.mediafire.com/view/?7045cz9au1ci7cd

A mountain bike has bad shock absorbers (w0/γ = 10) that oscillate with a period of 0.5 seconds after hitting a bump. If the mass of the bike and rider is 80kg, determine the value
of the spring constant k (remembering that there are two springs that share the mass equally).

The Attempt at a Solution



2pi/0.5 = w = 4pi But w = √(k/m)

16pi2 * 80 = k = 12633.

Since the two springs are in parallel we have the spring constant for an individual spring = 6317 N/m
 
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Might be OK but I can't read what it said in parentheses after " ...bad shock absorbers ...".
 
rude man said:
Might be OK but I can't read what it said in parentheses after " ...bad shock absorbers ...".

Its basically saying that the quality factor = 10

Could I have another opinion, please ?

Here's the entire question if you need it :

http://www.mediafire.com/view/?7045cz9au1ci7cd
 
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Couldn't open the link. But, no matter, you said Q = 10.

Which is a pretty high Q but still the fact that it's not infinite means your computation of k is off by a bit.

You first need to compute your natural frequency which = sqrt(k/m). Hint: it's not 2pi/0.5 rad/s. Close, but not equal to.

Use your knowledge of Q to get the natural frequency & then you're home free.
 
rude man said:
Couldn't open the link. But, no matter, you said Q = 10.

Which is a pretty high Q but still the fact that it's not infinite means your computation of k is off by a bit.

You first need to compute your natural frequency which = sqrt(k/m). Hint: it's not 2pi/0.5 rad/s. Close, but not equal to.

Use your knowledge of Q to get the natural frequency & then you're home free.

7045cz9au1ci7cdfg.jpg
 
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