Need help to solve an oscillation problem.

[Kkurenai]

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!

 

[Charles Link]

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.

 

[RPinPA]

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?

 

[Charles Link]

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 $\pi$.

 

[RPinPA]

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 $x = 0$.

 

[Kkurenai]

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.''

 

[haruspex]

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.