A problem on tuning fork.

  • #1
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Homework Statement


2 tuning forks A and B, when sounded together produce 4 beats. when B is loaded with wax, the beat frequency remains same. if frequency of A is 212 Hz, then frequency of B is?

Homework Equations


beat-frequency-formula.PNG


The Attempt at a Solution


since the beat frequency is 4 and frequency of A is 212 Hz, the frequency of B should be either 216 or 208. B is waxed which means it's frequency is decreased. if we assume the frequency of B as 216, on waxing the beat frequency decreases and considering the latter, waxing increases the beat frequency. but in the given question, it says beat frequency remains same.how could i solve this?
 

Answers and Replies

  • #2
DrClaude
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if we assume the frequency of B as 216, on waxing the beat frequency decreases
Will the beat frequency always decrease, however much wax is added?
 
  • #3
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Will the beat frequency always decrease, however much wax is added?
i guess so. doesn't it?
 
  • #4
DrClaude
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What happens if I add enough wax to get the frequency of B below 212 Hz?
 
  • #5
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What happens if I add enough wax to get the frequency of B below 212 Hz?
sorry.i don't get you
 
  • #6
DrClaude
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sorry.i don't get you
How does the beat frequency varies with the amount of wax as I add enough wax to get B vibrating at less than 212 Hz?
 
  • #7
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How does the beat frequency varies with the amount of wax as I add enough wax to get B vibrating at less than 212 Hz?
the beat frequency will go on increasing, if the frequency of B is constantly becomes lesser than 212
 
  • #8
DrClaude
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the beat frequency will go on increasing, if the frequency of B is constantly becomes lesser than 212
Correct. You should now be able to answer the problem.
 
  • #9
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Correct. You should now be able to answer the problem.
they have given that the beat frequency remains constant even after waxing, thats where i'm confused!
 
  • #10
DrClaude
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they have given that the beat frequency remains constant even after waxing, thats where i'm confused!
It is not that the beat frequency is constant, but the beat frequency with and without the wax is the same, for the particular amount of wax added.
 
  • #11
Gosh, it's hard to coach this kind of a question without just giving away the answer. Maybe just study the formula for the beat frequency a lot harder, and the answer will come.
 
  • #12
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It is not that the beat frequency is constant, but the beat frequency with and without the wax is the same, for the particular amount of wax added.
so the frequency of B is 208 Hz?
 
  • #15
DrClaude
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If B is 208 Hz without the wax, how will the beat frequency change when wax is added?
 
  • #16
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If B is 208 Hz without the wax, how will the beat frequency change when wax is added?
oh! if 208 Hz is waxed, then the beat frequency will be increased!
 
  • #17
DrClaude
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oh! if 208 Hz is waxed, then the beat frequency will be increased!
Yes. So what is the answer?
 
  • #19
DrClaude
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216Hz? is the answer?
What do you think? Does it explain the observed behavior of the tuning forks?
 
  • #20
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What do you think? Does it explain the observed behavior of the tuning forks?
yes cause if 216 Hz is frequencyof B without loading, it makes 4 beats with A. then on loading the frequency of B decreases to 208 Hz and in this case also it makes 4 beats with A. so 216 Hz should be the frequency of B before loading. right?
 
  • #21
DrClaude
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yes cause if 216 Hz is frequencyof B without loading, it makes 4 beats with A. then on loading the frequency of B decreases to 208 Hz and in this case also it makes 4 beats with A. so 216 Hz should be the frequency of B before loading. right?
Correct!
 
  • #23
I see this as a very interesting problem, even though in some sense it is trivial. I went through the same process, being confused (momentarily) by the wording of it, yet for myself, being pretty experienced with the beat frequency effect (from having been into ham radio), the answer came quickly. I had the experience that made me highly aware of the possibility of getting the same beat frequency from both sides, where one stationary frequency is combined with another one which is variable and sweeping across the point where both frequencies are the same.

However, I can certainly sympathise with how the question leads one to making an invalid presumption that is very hard to go back upon and question, when searching for insight into the problem. The broader question I have, is how can one learn, in the most general way, to recognize when one's own unquestioned presumption might be blocking insight into solving a problem. Incidentally, the wrong presumption here, was where the initial attempt to solve the problem, included the statement "if we assume the frequency of B as 216, on waxing[,] the beat frequency decreases..." . Actually that is not correct. Clearly true at first, but it omits to say that when the beat frequency reaches a certain point (zero in this case) it then increases again -- instead of continuing to decrease (into negative frequency land? :-) ! Of course the effect of absolute value brackets in a formula is what one has to be wary of, and yet, the effect of absolute value may so rarely come into play except in certain very special cases that it is easy to overlook.

I guess the answer is, when stuck, to remember to debug for all possible faulty presumptions. The major effort may be to first carefully identify all the presumptions. (Good luck on that..) Then one can try to think of all possible, if seemingly unlikely, variations of them, and work out the various results. Turn over enough rocks and likely there will be a bug under one of them.

And of course, one can specifically learn the rule of watching out for the implications of absolute value brackets... :-)
 

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