A string is stretched between fixed points separated by $75.0 ext{ cm}$. It is observed to have resonant fre

Question

A string is stretched between fixed points separated by $75.0 	ext{ cm}$. It is observed to have resonant frequencies of $420 	ext{ Hz}$ and $315 	ext{ Hz}$. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is:

Accepted Answer

1. **Identify Resonant Frequencies for a Fixed String:** For a string fixed at both ends, the resonant frequencies are integer multiples (harmonics) of the fundamental frequency, given by $f_n = n f_1$, where $n = 1, 2, 3, \dots$ and $f_1$ is the fundamental (lowest) resonant frequency .
2. **Set up the Equations:** The problem states there are no other resonant frequencies between $315 	ext{ Hz}$ and $420 	ext{ Hz}$, which means they are consecutive harmonics. Let them be the $n$-th and $(n+1)$-th harmonics:
   $$f_n = n f_1 = 315 	ext{ Hz}$$
   $$f_{n+1} = (n+1) f_1 = 420 	ext{ Hz}$$
3. **Calculate the Fundamental Frequency:** The difference between any two consecutive harmonic frequencies of a string fixed at both ends is equal to the fundamental frequency:
   $$f_{n+1} - f_n = (n+1)f_1 - n f_1 = f_1$$
   $$f_1 = 420 	ext{ Hz} - 315 	ext{ Hz} = 105 	ext{ Hz}$$
Therefore, the lowest resonant frequency (fundamental frequency) is $105 	ext{ Hz}$.

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