A wire of length L and linear density m is st | vedclass

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(B) The frequency of the $n^{th}$ harmonic for a string fixed at both ends is given by $f_n = n \cdot f_1$, where $f_1$ is the fundamental frequency.G

Vedclass · Accepted Answer

$4$

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(B) The frequency of the $n^{th}$ harmonic for a string fixed at both ends is given by $f_n = n \cdot f_1$, where $f_1$ is the fundamental frequency.Given that the wire resonates at $320 	ext{ Hz}$ in the $P^{th}$ harmonic, we have $f_P = P \cdot f_1 = 320 	ext{ Hz}$.The next higher frequency in a successive mode is the $(P+1)^{th}$ harmonic, given as $f_{P+1} = (P+1) \cdot f_1 = 400 	ext{ Hz}$.Subtracting the first equation from the second: $(P+1)f_1 - P f_1 = 400 - 320$.This gives $f_1 = 80 	ext{ Hz}$.Substituting $f_1$ back into the first equation: $P \cdot 80 = 320$.Therefore, $P = 320 / 80 = 4$.

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