Need help to solve an oscillation problem.

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Homework Statement


It's an oscillation problem. I have to find the FIRST time when the spring-mass system will have E (mechanical energy) = K (kinetic energy), if x(t)=12sin(5t+3,5). (t is time in seconds, x is lenghtof the system in cm).

Homework Equations


E=1/2kA^2
K= 1/2kx^2

The Attempt at a Solution


What I think: If E = K, there is no potential energy (U). So we are searching the first time when K will be maximal. I know that Kmax in a simple harmonic motion is when V is max. So, in the trigonometry circle : at π and 2π, 3, 4 etc. Since Φ start at 3,5, little after π, i think that the first time will be at 2π rad (where K is max). But when i use 2π in the equation [ x(t)=12sin(5t+3,5)], i come up with a negative time...
Thanks for helping me sorting this out!
 

Answers and Replies

  • #2
Charles Link
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Just to clarify, ## x(t)=12 \sin(5t+3.5) ## if I interpreted it correctly. (Europe sometimes uses a comma for the decimal point). ## \\ ## Do they want the first time that it occurs where ## t>0 ## ? The statement of the problem seems somewhat unclear.
 
  • #3
RPinPA
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Your reasoning is fine. So it will be all kinetic energy at ##2\pi##, which means also at 0, ##\pm 2\pi##, ##\pm 4\pi##, etc. Nothing wrong with your reasoning except for your assumption that ##2\pi## had to be the first without examining the argument.

So find the first number of that form that gives a positive ##t##. It's not ##2\pi##, what's the next time where the function crosses 0?
 
  • #4
Charles Link
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Your reasoning is fine. So it will be all kinetic energy at ##2\pi##, which means also at 0, ##\pm 2\pi##, ##\pm 4\pi##, etc. Nothing wrong with your reasoning except for your assumption that ##2\pi## had to be the first without examining the argument.

So find the first number of that form that gives a positive ##t##. It's not ##2\pi##, what's the next time where the function crosses 0?
Don't forget the odd multiples of ## \pi ##.
 
  • #5
RPinPA
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Don't forget the odd multiples of ## \pi ##.
Oops, yes, you're right.

The simpler reasoning is this: The mechanical energy is all kinetic when the potential energy is 0. Since PE = ##(1/2)kx^2## then this happens when ##x = 0##. So you're looking for a ##t## value when ##x = 0##.
 
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  • #6
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RPinPA
Just to clarify, ## x(t)=12 \sin(5t+3.5) ## if I interpreted it correctly. (Europe sometimes uses a comma for the decimal point). ## \\ ## Do they want the first time that it occurs where ## t>0 ## ? The statement of the problem seems somewhat unclear.
Thanks Charles,
Yes, it's Φ = 3,5. The question is ''Find the 1st time when E=K.''
 
  • #7
haruspex
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when i use 2π in the equation [ x(t)=12sin(5t+3,5)], i come up with a negative time...
So try the next one.
 

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