Help Explain. Physics problem

[prettyNURSEtoMD]

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A 1.00kg mass oscillates according to the equation x=0.600cos8.20t, where x is in meters and t is in seconds.

Determine the kinetic and potential energy when x=0.350m

I've gotten the amplitude, frequency, and total energy correct, but I'm really stuck on these two.

I've done every formula I know and I'm still not getting the right answer.

E = 1/2kA2

PE = 1/2kx2

KE = E - PE

Can someone help me? Thanks

 

[DrknoSDN]

That looks like a hideous question that is not going to be on an MCAT, I have no idea how you could figure that out without a calculator.. and you don't get one during the test i believe.

Looks more like a physics textbook problem.

 

[siffster]

Ok so... The equation for the displacement of an oscillating spring is x=Acos(wt). As you can see, A(amplitude) corresponds to 0.6 in the equation given, and w corresponds to 8.20.
The energy of the system at a specific location is: Total energy : KE + PE(spring)= 1/2mv^2 + 1/2kx^2. if youre at max amplitude, it would all be potential energy(1/2kA^2).

So you want to know the energy at x=0.350, but you're missing the velocity and the k-constant. as you might remember, you can find the k-constant through the relationship with w(angular velocity): w= sqrt(k/m). so k=mw^2. the angular velocity was given in the equation mentioned=8.2, so k=(1)(8.2)^2, which im gonna approximate to 65.

So you mentioned that the kinetic energy is KE = E - PE. Assuming the spring isn't damped, the total energy is E=1/2kA^2. So E=1/2(65)(0.6)^2 =

solve for the potential energy at x= 0.35 ----> 1/2(65)(0.35)^2 and substract the result from the total energy: KE=E-PE.