# 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

Staff Emeritus
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.