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Fundamental Frequency!!! |
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| Jun20-09, 04:31 AM | #1 |
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Fundamental Frequency!!!
A nylon string is stretched between fixed supports 0.75m apart. Experimental plucking of the string shows that several standing waves can exist on the string. Two such standing waves have frequencies of 225Hz and 300Hz with no other frequencies in between.
Q1. What is the lowest frequency (fundamental) frequency that can exist on the string? I tried to use f=v/2L = Root(t/m/L)/2L But the tension and mass was not given. I'm pretty sure that would have something to do with those frequencies provided. But I have no clue of how to do it. Thanks ^^ |
| Jun20-09, 04:35 AM | #2 |
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HINT: How is the frequency of the nth harmonic related to the fundamental mode? |
| Jun20-09, 04:40 AM | #3 |
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I'm not sure very sure... |
| Jun20-09, 04:47 AM | #4 |
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Fundamental Frequency!!!Can you use this expression together with the information provided to create a system of two equations? |
| Jun20-09, 04:56 AM | #5 |
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Throught my calculations.... So f1=300-225? |
| Jun20-09, 04:57 AM | #6 |
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| Jun20-09, 04:58 AM | #7 |
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Ohh ! I got it!!!
Because the Nth frequency is the n*fundamental frequency! Is that right? |
| Jun20-09, 05:00 AM | #8 |
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Thank u !!! :D
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| Jun20-09, 05:38 AM | #9 |
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Okay, exactly what answer did you get? It does NOT follow that because [itex]225= n*f_1[/itex], that 330 must be the next harmonic. [itex]330= m*f_1[/itex], certainly, but m is not necessarily n+1.
330- 225= 105. But 105 is NOT a divisor of 330: that is, 330 is not equal to n*105 for any integer n so 105 is NOT the "fundamental frequency". What is true is that [itex]f_n= n*f_1[/itex] so the fundamental frequence must be a factor of both 330 and 225 (and, so, 105). 330= 3*110= 3*5*22= 2*3*5*11. 225= 5*45= 5*5*9= 32*52. What is the largest number that divides both of those? (Even then there is no guarentee that a fundamental frequency is the greatest common factor. There four possible fundamental frequencies for this. |
| Jun20-09, 08:08 AM | #10 |
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