How Do You Calculate Spring Stretch When Kinetic and Potential Energy Are Equal?

[quickclick330]

One end of a massless spring is fixed and the other end is attatched to a mass. The mass slides on a horizontal frictionless table. The spring is parallel to the table. The mass-spring system oscillates with an amplitude of 4 cm. What is the magnitude of stretch of the spring when the kinetic energy of the mass is equal to the spring potential energy at that instant? (The spring ptotential is defined to be zero when at its relaxed length.)

I have:

KE = U

1/2*m*v^2 = 1/2ksX^s

My question is, how do I use the amplitude for this equation to work?? thanks! :-)

 

[Doc Al]

Hint: The total energy at any point equals the sum of KE + PE.

 

[quickclick330]

I know i need to use one instance of time to solve for this instant in time for energy, but I'm unsure about which points to choose?

 

[Doc Al]

What's the total energy expressed in terms of the amplitude?

 

[quickclick330]

TE = U + KE
= 1/2ksX^2 + 1/2mv^2
when the amplitude = 4, then velocity = 0?
= 1/2*ks4^2 + 0
E = 8ks ?

 

[Doc Al]

Good. You're on the right track. Rather than use numbers, let's just call the amplitude A:
E = 1/2kA^2.

So, at the point where KE = PE, how much are each in terms of the total energy?

 

[quickclick330]

i think I'm confused. I keep wanting to use PE as 1/2kA^2 and KE 1/2mv^2 but that doesn't really help me much?

 

[Doc Al]

The PE at any point is 1/2kX^2. At the point we want, how much of the total energy is PE? Write that as an equation.

 

[quickclick330]

Since total energy = PE + KE

and PE =KE

so that means that TE has to equal 2*PE

E = 2(1/2kA^2)

is this right?

 

[Doc Al]

Since total energy = PE + KE

and PE =KE

so that means that TE has to equal 2*PE

Yes!

E = 2(1/2kA^2)

Almost. The left side (E) is the total energy, so plug in the total energy in terms of A. On the right hand side, write the PE in terms of X. (Then just solve for X.)

 

[quickclick330]

ahh. that makes sense. thanks for your help!