Simple Harmonic Motion: Oscillator

AI Thread Summary
An oscillator with a mass of 600 g and a period of 0.50 s experiences a 2.0% decrease in amplitude with each complete oscillation, starting from an initial amplitude of 6 cm. The discussion emphasizes that after each oscillation, the amplitude is multiplied by 0.98, leading to a compound decrease. Some participants clarify that the decrease is indeed based on the previous amplitude, resembling a reverse compound interest scenario. The calculation for the amplitude after 25 oscillations involves repeated multiplication by 0.98. Ultimately, the conversation highlights the importance of understanding the nature of the amplitude decrease in oscillatory motion.
TJC747
Messages
5
Reaction score
0
An oscillator with a mass of 600 g and a period of 0.50 s has an amplitude that decreased by 2.0% during each complete oscillation. If the initial amplitude is 6 cm, what will be the amplitude after 25 oscillations?

I should most likely be using T = 2pi*sqrt L/g, as well as the many derivations of Hooke's laws, yet I cannot piece them all together. Help would be appreciated. Thanks.
 
Physics news on Phys.org
You don't need to use any formulas. Just think. After one oscillation, amplitude = 0.98 * initial amplitude, correct? After two oscillations, A=0.98*0.98*Ai. After three, 0.98*0.98*0.98*Ai. After 25?
 
ideasrule said:
You don't need to use any formulas. Just think. After one oscillation, amplitude = 0.98 * initial amplitude, correct? After two oscillations, A=0.98*0.98*Ai. After three, 0.98*0.98*0.98*Ai. After 25?

Are you sure it's a constant decrease of .02? Wouldn't it be a decrease by 2.0% of the previous complete oscillation, so the amplitude as

1.00 - 1.00*.02
.98 - .98*.02
.9604 - .9604*.02
.9401 - .9401*.02

and so on.

This looks rather like compound interest but in "reverse".
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top