Two tuning forks A and B sounded together giv | vedclass

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(C) Let the frequency of the first fork be $f_1$ and that of the second be $f_2$.For a resonance tube closed at one end,the fundamental frequency is g

Vedclass · Accepted Answer

$150 \, Hz, 144 \, Hz$

Vedclass · Answer

(C) Let the frequency of the first fork be $f_1$ and that of the second be $f_2$.For a resonance tube closed at one end,the fundamental frequency is given by $f = \frac{v}{4L}$,where $v$ is the speed of sound and $L$ is the length of the air column.Thus,$f_1 = \frac{v}{4 	imes 24}$ and $f_2 = \frac{v}{4 	imes 25}$.Since $f_1 > f_2$,we have the ratio $\frac{f_1}{f_2} = \frac{25}{24}$.Given that the beat frequency is $6 \, Hz$,we have $f_1 - f_2 = 6$.Substituting $f_1 = f_2 	imes \frac{25}{24}$ into the beat equation: $f_2 	imes \frac{25}{24} - f_2 = 6$.$f_2 	imes (\frac{25-24}{24}) = 6 \implies f_2 = 6 	imes 24 = 144 \, Hz$.Then,$f_1 = 144 + 6 = 150 \, Hz$.Therefore,the frequencies are $150 \, Hz$ and $144 \, Hz$.

Two tuning forks $A$ and $B$ sounded together give $6$ beats per second. With an air resonance tube closed at one end,the two forks give resonance when the two air columns are $24 \, cm$ and $25 \, cm$ respectively. Calculate the frequencies of the forks.

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