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Fundamental Frequency

  • #1
Which of the following could be the fundamental freq. for a vibration that has an overtone freq. of 990 Hz?

a. 330
b. 660
c. 148
d. 1980
e. 1990
(All in Hz.)

The formula that I think you have to use is f_n = n*f_1, where f_1 is fud. freq.

Then, I thought

f_2 = 2*f_1, where f_2 is the overtone freq. (is these a correct assumption?)

but 990/2 is 495.

Could someone direct me to the correct path of getting the solution?

I'm confused over the definitions of overtone and fund. freq.

Thanks.
 
Last edited:

Answers and Replies

  • #2
Gokul43201
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Science Advisor
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17
There's no reason to assume than n=2, so you just leave it as n.

So you have [itex]f_n = nf_1 [/itex]
Or [itex]f_1 = \frac {990}{n} [/itex]

Which of the choices satisfies this condition for some integer n ?

Can you say if [itex]f_1[/itex] will be greater than or less than 990 Hz, from this ? Does this rule out any choices ?

The alternative is simply to notice that [itex]f_n/f_1 = n [/itex], so try the remaining choices of [itex]f_1[/itex] to see which one gives you an integer ratio.
 
Last edited:
  • #3
Is the answer A?

The fud freq. would be less than 990, would it not? That leaves A and B, and 990/3 is 330.
 
  • #4
Gokul43201
Staff Emeritus
Science Advisor
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That's right.
 

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