Question

Q night... Help me.. Simple harmonic

Answer 1

No 3

Answer 2

The total energy of this oscillator would be conserved (that is, the total energy is the same no matter what the position and velocity are) . When an object comes to a stop it has no kinetic energy. Do you know the potential energy of a simple harmonic oscillator?

Answer 3

Ok.. 1/2kA^2

Answer 4

Can i use simple equation of x=a cos(kx+wt)

Answer 5

That's not the right equation for a simple harmonic oscillator. Note that you have x on both sides of your equation! So it is not a solution for x in terms of t!
The motion of a simple harmonic oscillator is described by
\[x = Acos(\omega t + \theta)\]

But for this problem, you just need to find A, which is the maximum value of x.

If the oscillator is at position x, what is its potential energy at that moment?

Answer 6

Potential energy become 0 ???

Answer 7

Potential.... 1/2 ka^2

Answer 8

But how to get k.. .???

Answer 9

If x is 0, then there is no energy stored in the spring. As you increase x, you increase the stored potential energy. So the equation for potential energy at position x should have an x in it!

You can solve for k in terms of the mass. You can solve this problem without knowing what the mass is.

\[F = ma = -kx\]

Answer 10

Can i subtitute.. K/m = w^2... Or it will be more complex..

Answer 11

So it will be 1/2(ma(A^2)/x))=????

Answer 12

Does that make sense? Should the potential energy get smaller or larger as the distance from the equilibrium increases?

Yes, you can substitute k/m = ω^2. Even though there is not enough information to find k, you can find ω^2 using the initial position and acceleration.