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Frequencies & Waves

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Homework Statement


1)A stretched wire vibrates in its first normal mode at a frequency of 383 Hz. What would be the fundamental frequency if the wire were one fourth as long, its diameter were quadrupled, and its tension were increased four-fold?

2)A 0.0125 kg, 1.88 m long wire is fixed at both ends and vibrates in its simplest mode under a tension of 218 N. When a tuning fork is placed near the wire, a beat frequency of 5.10 Hz is heard. The beat frequency increase as the the tension of the wire decreases; what is the frequency of the tuning fork?


Homework Equations


For question 1:
f = 1/2L ( sqrt( T/mu)




The Attempt at a Solution


I really dont know how to start in these two questions, given with a diameter in part 1, but I know the formula to use. And I'm not sure what equation Im going to use for question 2.

Thanks
 

Answers and Replies

  • #2
Redbelly98
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Homework Equations


For question 1:
f = 1/2L ( sqrt( T/mu)
Okay, can you answer this: what would be the fundamental frequency if the wire were one fourth as long, and nothing else changed?



The Attempt at a Solution


I really dont know how to start in these two questions, given with a diameter in part 1, but I know the formula to use. And I'm not sure what equation Im going to use for question 2.
Your textbook or class lecture notes should have a discussion of beat frequencies, along with an equation or two.
 
  • #3
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In part one, if only the length is changed, the equation would become

f = [tex]2/L[/tex]([tex]\sqrt{T/mu}[/tex])

In part two, I checked my notes and saw f[beat] = f2- f1
I have f1 but I dont have f2. So I guess I should use v^2*mu = T? then what should I do next? or is there another equation that relates tension with frequency?

Thanks
 
  • #4
Redbelly98
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Question #1

In part one, if only the length is changed, the equation would become

f = [tex]2/L[/tex]([tex]\sqrt{T/mu}[/tex])
Okay, so in that case the fundamental frequency would be ____ Hz?

.

Question #2

In part two, I checked my notes and saw f[beat] = f2- f1
Yes, good, though actually there should be an absolute value sign there:
f[beat] = |f2- f1|
I have f1 but I dont have f2.
As far as I can tell, you don't really have f1, do you? fbeat was given, but not f1 or f2.

So I guess I should use v^2*mu = T? then what should I do next?
Well, how exactly do you plan to use that equation? What new information will you get from using it?

or is there another equation that relates tension with frequency?
Not that I'm aware of.
 
  • #5
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But in question 1, I'm not given with tension or mu. Is it possible to get the f?

In question 2:

I thought I would be able to get lambda with string speed, but when I check the other equation again to get lambda I would need wave speed not string speed...So I guess the equation that I mentioned was wrong.
 
  • #6
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In question 1 again, Do I need to find the length first using the frequency while ignoring mu and tension? I also know the length of the in the first frequency is the same as the length in the 2nd frequency when the length becomes one fourth as long.


Thanks
 
  • #7
Redbelly98
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#1:

You do not need to know the length, tension, or mu.

In the problem statement we are told that (1/2L) · (sqrt( T/mu)) is equal to ____(?)

If the length is one fourth as long, then the new frequency (2/L)·( sqrt( T/mu)) would be ____(?) times as much.

Fill in those blanks.
 
  • #8
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#1:

You do not need to know the length, tension, or mu.

In the problem statement we are told that (1/2L) · (sqrt( T/mu)) is equal to ____(?)

If the length is one fourth as long, then the new frequency (2/L)·( sqrt( T/mu)) would be ____(?) times as much.

Fill in those blanks.
First blank: 383 Hz

Second blank: 1532 Hz. 4 times as much
 
  • #9
Redbelly98
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First blank: 383 Hz

Second blank: 1532 Hz. 4 times as much
Good, it looks like you have the idea. Can you apply that same reasoning to

1. increasing the tension 4-fold?
2. quadrupling the diameter (think about how that would affect mu)?
 
  • #10
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If the tension is four-fold, then the new frequency would be 766 Hz ( twice as much ).

If the diameter is quadrupled, then the new frequency would be 191.5 Hz ( half as much ).

EDIT: I got the answer! its 766 Hz thanks :D.

Question 2:

f[beat] = | f2 - f1 |

I didnt know that 5.10 Hz was the beat frequency. So I guess I wont use the formula using speed. How should I start with this question?

Thanks again
 
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  • #11
Redbelly98
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We know that fbeat = |fwire - ffork|

We want ffork. So we need to know the values of the other terms in the equation in order to get ffork.
 

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