SHM of a Tuning Fork: Calculating Maximum Velocity and Acceleration

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Homework Help Overview

The discussion revolves around a problem related to simple harmonic motion (SHM) involving a tuning fork. Participants are tasked with calculating the maximum velocity and maximum acceleration of the prong based on its frequency and the distance it moves between extreme positions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants share their calculations for maximum velocity and acceleration, questioning the correctness of their results. There is an emphasis on the importance of understanding amplitude and its definition in the context of SHM.

Discussion Status

Some participants have provided calculations and are seeking validation of their results. Others have pointed out potential errors regarding the definition of amplitude and its application in the calculations. The discussion is ongoing, with participants exploring different interpretations and seeking clarification.

Contextual Notes

There is a noted confusion regarding the amplitude of motion, with participants discussing the difference between the total distance moved and the maximum displacement from the equilibrium position. This indicates a need for careful consideration of definitions in the problem context.

shorti2406
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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 :smile:

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!
 
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shorti2406 said:
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 :smile:

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!

Not quite- What is the amplitude of the motion?

Plus- you will need to be a lot more careful with the units
 
Last edited:
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?
 
The amplitude is still not correct.
 
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!
 
shorti2406 said:
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!
You'll smack yourself in the head when you realize it:
The distance the prong moves between its extreme positions is 2.29 mm.

Remember that amplitude is the maximum displacement from the equilibrium position, not the distance between the extreme positions.
 
You're right. I have no idea why I didn't realize that! Thank you so much for you help.
 

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