Tuning fork problem

  • Thread starter lollypop
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  • #1
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hi:
my questions says the following:
A tuning fork labeled 392 Hz has the tip of each of its two prongs vibrating with an amplitude of 0.600 mm.
What is the maximum speed of the tip of a prong?

for this part i found omega= frequency*2pi
then plugged that in V= omega*Amplitude, so my answer is 1.48 m/s.

Now in the second part of the problem they ask:
A housefly with mass 0.0270 g is holding on to the tip of one of the prongs. As the prong vibrates, what is the fly's maximum kinetic energy? Assume that the fly's mass has a negligible effect on the frequency of oscillation.

is this the formula i have to use ?--> E=.5mv^2 + .5 kx^2, if it is, please how can i find x ? do i get k from omega = sqrt(k/m). i'm not sure what to do for this part. which equation should i use?
:confused:
 

Answers and Replies

  • #2
Why would you include the potential energy of the fork? That's got nothing to do with the fly's kinetic energy.

cookiemonster
 
  • #3
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i don't know which one to use, using just .5mv^2 gives me the wrong answer, is there any other formula i may use???
 
  • #4
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The idea is that the velocity of the housefly is the same as the velocity of the tuning fork, and therefore the kinetic energy of the housefly is proportional to that of the fork. When does the fork have the maximum kinetic energy? Can you find its velocity at that point?
 
  • #5
HallsofIvy
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lollypop said:
i don't know which one to use, using just .5mv^2 gives me the wrong answer, is there any other formula i may use???

You have the speed correct. Did you convert the flies mass to kg. so that you get the kinetic energy in Joules? (Or convert speed to cm/s and use the mass as given in grams to get the kinetic energy in ergs.)
 
  • #6
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What is the formula for acceration? For velocity is omega * amplitude
 

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