Sound waves question - vibrations and amplitudes

AI Thread Summary
The discussion focuses on calculating the frequency, period, and distance traveled by a tuning fork vibrating at 235 times in 3.1 seconds with an amplitude of 0.59 mm. The frequency is determined to be approximately 75.81 Hz, and the period is corrected to 0.013 seconds. For the distance traveled in one second, it is clarified that the total distance is four times the amplitude multiplied by the frequency, resulting in approximately 178.9 mm. The conversation emphasizes the distinction between displacement and distance traveled during oscillation. Overall, the calculations and concepts of wave motion are thoroughly analyzed and clarified.
FlorenceC
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A tuning fork vibrates 235 times in 3.1 seconds with an amplitude of 0.59mm. Determine..
i) its frequency
ii) its period
iii) the distance the end of one of the fork tines travels in one second

for i) The answer I got is f=75.81Hz
ii) The answer I got is 0.13s
iii) ?
 
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hz is 1/s so 235/3.1s

period is 1/frequency s
 
You may want to check your calculation for part (ii) again! I think it is 0.013s
 
I think a little more info is needed for part (iii) because you don't know where it is starting from. Do we want to find the distance it travels after 1 second from it's equilibrium position or from its maximum displacement?
 
ofeyrpf said:
I think a little more info is needed for part (iii) because you don't know where it is starting from. Do we want to find the distance it travels after 1 second from it's equilibrium position or from its maximum displacement?

I'm going to guess equilibrium, since this is only gr. 11 physics. The supply teacher said there are 4 amplitudes in 1 vibration, so the distance is 4*0.59*frequency. I tried it, but I got an unreasonable answer and I don't think there are 4 amplitudes in a vibration.
 
FlorenceC said:
I'm going to guess equilibrium,
Another reasonable guess is that it means on average (which is fact what you computed).
The supply teacher said there are 4 amplitudes in 1 vibration, so the distance is 4*0.59*frequency. I tried it, but I got an unreasonable answer and I don't think there are 4 amplitudes in a vibration.
Starting from equilibrium, how far will it travel in the first quarter period? How far in the second quarter?
What answer did you get?
 
haruspex said:
Another reasonable guess is that it means on average (which is fact what you computed).
Starting from equilibrium, how far will it travel in the first quarter period? How far in the second quarter?
What answer did you get?

First quarter = 44.73 mm
Second quarter = 89.4 6mm
 
FlorenceC said:
First quarter = 44.73 mm
Second quarter = 89.4 6mm
How did you get those numbers? The amplitude is only 0.59 mm.
 
haruspex said:
How did you get those numbers? The amplitude is only 0.59 mm.

For the first quarter, there is 1 amplitude, so (1*0.00059*75.81)
For the second quarter, there are 2 amplitude, so (2*0.00059*75.81)

I multiplied by the frequency to get the answer for 1 second.
 
  • #10
Displacement is not the same as the distance travelled. If you go to school 500 m far, you walk 500 m and your displacement is also 500 m. If you go there and back, your displacement is zero, but you walked 1000 m, your "distance travelled" or "distance covered" is 1000 m. The tip of the fork travels from equilibrium to maximum displacement (amplitude) during a quarter of the period. Then it moves backwards, reaching zero displacement in quarter period again, but traveling A distance again, so the distance traveled in half period is 2A. The whole distance traveled in one period is 4A. The number of periods in one second is equal to the frequency. So the total distance traveled in one second is? ehild
 

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  • #11
ehild said:
Displacement is not the same as the distance travelled. If you go to school 500 m far, you walk 500 m and your displacement is also 500 m. If you go there and back, your displacement is zero, but you walked 1000 m, your "distance travelled" or "distance covered" is 1000 m. The tip of the fork travels from equilibrium to maximum displacement (amplitude) during a quarter of the period. Then it moves backwards, reaching zero displacement in quarter period again, but traveling A distance again, so the distance traveled in half period is 2A. The whole distance traveled in one period is 4A. The number of periods in one second is equal to the frequency. So the total distance traveled in one second is?


ehild

so the answer is 4*0.00059*75.81 = .1789m or 178.9mm
 
  • #12
FlorenceC said:
so the answer is 4*0.00059*75.81 = .1789m or 178.9mm

Yes it is, but round it to three significant digits.

ehild
 
  • #13
FlorenceC said:
For the first quarter, there is 1 amplitude, so (1*0.00059*75.81)
For the second quarter, there are 2 amplitude, so (2*0.00059*75.81)

I multiplied by the frequency to get the answer for 1 second.
I had asked how far in one quarter period, i.e. one quarter of a complete oscillation.
You had expressed doubts that it traveled four amplitudes in each period; I was trying to get you to see that it does.
 
  • #14
ehild said:
Yes it is, but round it to three significant digits.

ehild

haruspex said:
I had asked how far in one quarter period, i.e. one quarter of a complete oscillation.
You had expressed doubts that it traveled four amplitudes in each period; I was trying to get you to see that it does.

I finally understand it now. Thank you so much ehild and haruspex.
 
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