# Homework Help: SHM Question

1. May 5, 2005

### shorti2406

Q: The prong of a tuning fork moves back and forth when it is set into vibration. The distance the prong moves between its extreme positions is 2.29 mm. If the frequency of the tuning fork is 440.2 Hz, what are the maximum velocity and the maximum acceleration of the prong? Assume SHM.

I was just hoping somebody could tell me if I did this problem correctly

For Vmax I have : (2л) x (440.2Hz) x (0.0029cm) = 8.02 m/s

For amax I have : [(2л x 440.2Hz)^2] x (0.0029cm) = 22185 m/s2

I would appreciate any input! Thanks!

2. May 5, 2005

### OlderDan

Not quite- What is the amplitude of the motion?

Plus- you will need to be a lot more careful with the units

Last edited: May 5, 2005
3. May 5, 2005

### shorti2406

Oh, okay..... so then:

Vmax : (2л) x (440.2Hz) x (0.00229m) = 6.33 m/s

amax : [(2л x 440.2Hz)^2] x (0.00229m) = 17518 m/s2

Does that look about right? Or am I still way off?

4. May 5, 2005

### Staff: Mentor

The amplitude is still not correct.

5. May 5, 2005

### shorti2406

Well, I think I'm just stuck then. Can anybody give me a clue as to what the correct amplitude shoudl be? I would really appreciate it!

6. May 5, 2005

### Staff: Mentor

You'll smack yourself in the head when you realize it:
Remember that amplitude is the maximum displacement from the equilibrium position, not the distance between the extreme positions.

7. May 5, 2005

### shorti2406

You're right. I have no idea why I didn't realize that! Thank you so much for you help.

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