# Damped harmonic motion

1. Jun 9, 2009

### nokia8650

http://img13.imageshack.us/img13/9091/53337497.th.jpg [Broken]

Can someone please help me with the problem above? I am unable to start it. Clearly, using the constant acceleration "suvat" equations, 0.5ft^2 is the distance obtainined, however I am unable to proceed.

Last edited by a moderator: May 4, 2017
2. Jun 9, 2009

### HallsofIvy

So you are able to do part (a), then. At any moment, t, the force on the particle is man2 times the extension of the spring. That is y and, using (a), $y=(1/2)ft^2-x$.

$$m\frac{d^2x}{dt^2}= man^2((1/2)ft^2- x)$$

Are you sure about "modulus of elasticity man2"? That "a" doesn't seem to fit.

Last edited by a moderator: May 4, 2017
3. Jun 9, 2009

### nokia8650

For the tension, you forgot to divide by the natural length = a, and hence the a seems to fit! Thanks alot for the help though, it makes sense now! How would I do the last part of finding the maximum tension in the spring?

Thanks

Last edited: Jun 9, 2009
4. Jun 12, 2009

### nokia8650

Thanks

5. Jun 12, 2009

### Cyosis

The tension is maximum when the string's extension is at a maximum. You have an expression for the displacement. How do you find the maximum of this function?

6. Jun 13, 2009

### nokia8650

Would I say y = 0.5ft^2 - (particular solution of the equation), and then find when dy/dt = 0, to find the time at which y is a max, and then subv this time back in, to find ymax, and then use the tension equation?

Thanks

7. Jun 13, 2009

### Cyosis

Sounds like a good idea.