Simple harmonic motion and energy - yet another answer key disagreement?

[clairez93]

Homework Statement

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is xm.
When it is at x = x1, its kinetic energy is K = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = –1/2xm, the kinetic and potential energies are:

A) K = 5J and U = 3J
B) K = 5J and U = –3J
C) K = 8J and U = 0
D) K = 0 and U = 8J
E) K = 0 and U = –8J

Homework Equations

<br /> E = 1/2kA^2<br />

1/2kA^2 = 1/2mv^2 + 1/2kx^2

The Attempt at a Solution

<br /> 1/2kA^2 = 8

1/2mv^2 + 1/2kx^2^ = 8

A = 4/\sqrt{k}

x = -1/2(4/\sqrt{k}) = -2/\sqrt{k}

1/2mv^2 + 1/2k(-2/\sqrt{k})^2^ = 8

1/2mv^2 + 2 = 8

K = 6 J

U = 8 - 6 = 2 J<br /> <br />


As you see, this is not one of the choices. Am I doing something wrong?

 

[clairez93]

Sorry about the bad formatting earlier; I have fixed it now.

 

[clairez93]

Looking at it, I guess by process of elimination it has to A because energy isn't negative, thus B, C, and E are ruled out. And it cannot have full kinetic energy between equilibrium and amplitude, so D is ruled out.

However, it disagrees with my calculations.

 

[LowlyPion]

Don't you know what the total energy in the system is at x₁?

Don't you also know at x = 0 is where U = 0? And also at X_m is where K = 0, so ... which answer meets these requirements?

Edit: Ooops. Looks like you figured it out.

 

[gabbagabbahey]

It looks to me like your original solution is fine, and the answer key is incorrect...

The total energy is 8J, so \frac{1}{2}kA^2=8 \, \text{J}. And so at x=\frac{-x_m}{2}=\frac{-A}{2} , the potential energy is:

U(x)=\frac{1}{2}kx^2=\frac{1}{2}k \left( \frac{-A}{2} \right)^2=\frac{1}{4} \left( \frac{1}{2}kA^2 \right)= \frac{1}{4}(8 \, \text{J})=2 \, \text{J}