UPSB v4
Off-topic / Physics help. (Damped Oscillations...)
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Date: Thu, Jan 20 2011 16:53:36
Been working on this all last night, basically I know it's simple but I don't have school today so I can't ask the teacher, so if someone could help me out... [B]While seated on a tall bench, extend your lower leg a small amount and then let it swing freely about your knee joint, with no muscular engagement. It will oscillate as a damped pendulum. The figure is a graph of the lower leg angle versus time in such an experiment. Estimate the time constant for this oscillation. tau = ___ s[/B] I know tau is the inverse of the Decay Constant, can't find any equations / don't know ho to manipulate the variables to estimate however. ____________-- [B]The figure shows two springs, each with spring constant 18 N/m, connecting a 2.1 kg block to two walls. The block slides on a frictionless surface. If the block is displaced from equilibrium, it will undergo simple harmonic motion.[/B] What is the frequency of that motion? f= _____ Hz ----------- @strat1227;
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Date: Thu, Jan 20 2011 19:15:47
τ is the time it takes the amplitude to reach 1/e of it's original value (1/e ~ 0.36) spring constants are additive, so for this normally it'd be w = sqrt(k/m) this time it's w=sprt((k1+k2)/m) @shoeman6
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Date: Thu, Jan 20 2011 19:36:33
@strat1227; using f0 = [(1/2π) x (√(k/m)] I get 6.5 Hz, which is incorrect, does this equation not apply, or am I calculating it incorrectly? k = (k1+k2)
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Date: Thu, Jan 20 2011 19:57:43
@shoeman6 http://scienceworld.wolfram.com/physics/SpringsTwoSpringsandaMass.html
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Date: Thu, Jan 20 2011 19:59:12
@shoeman6 i think the problem is it that you did (1/2)*pi and not 1/(2pi)
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Date: Thu, Jan 20 2011 20:02:13
@strat1227; still inccorect ( I got 4.14) , is the period affected, should 4.14 be multiplied by a degree of pi? :\ idk what's wrong
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Date: Thu, Jan 20 2011 20:03:30
YAY! STRAT YOU FIXED IT @strat1227; <3 thanks. :]
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Date: Thu, Jan 20 2011 20:04:15
http://www.google.com/search?sourceid=chrome&ie=UTF-8&q=sqrt(36/2.1)#sclient=psy&hl=en&safe=off&q=sqrt(36%2F2.1)%2F(2*pi)&aq=f&aqi=&aql=&oq=&pbx=1&fp=25a0b2344dc0e416 @shoeman6 dunno what you're saying, just do sqrt(k/m)/(2pi) ...
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Date: Sun, Jan 23 2011 02:28:02
@strat1227; A 2.0 kg block oscillates up and down on a spring with spring constant 210N/m . Its initial amplitude is 14 cm. If the time constant for damping of the oscillation is 4.0s , how much mechanical energy has been dissipated from the block-spring system after 12 s? |deltaE|= ____ J
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Date: Sun, Jan 23 2011 02:31:40
lol come on this one is easy you shouldn't need my help i showed you last time what the time constant means, so you use that to find its amplitude after 12 seconds, you find that energy and subtract it from initial energy @shoeman6