(A) For a pipe closed at one end,the fundamental frequency is given by $n = \frac{v}{4L}$.For tuning fork $A$,$n_A = \frac{v}{4 \times 37.5 \times 10^

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(A) For a pipe closed at one end,the fundamental frequency is given by $n = \frac{v}{4L}$.For tuning fork $A$,$n_A = \frac{v}{4 \times 37.5 \times 10^{-2}}$.For tuning fork $B$,$n_B = \frac{v}{4 \times 38.5 \times 10^{-2}}$.Since $n_A > n_B$ (as $L_A $\frac{v}{4 \times 0.375} - \frac{v}{4 \times 0.385} = 8$.$\frac{v}{4} \left( \frac{38.5 - 37.5}{37.5 \times 38.5} \times 100 \right) = 8$.$\frac{v}{4} \left( \frac{1}{1443.75} \times 100 \right) = 8 \implies v = \frac{8 \times 4 \times 1443.75}{100} = 462 \, m/s$.Now,$n_A = \frac{462}{4 \times 0.375} = 308 \, Hz$.$n_B = n_A - 8 = 300 \, Hz$.

Class 11 Physics Waves and Sound Beats and Tuning fork

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Two tuning forks $A$ and $B$ produce $8 \, Hz$ beats per second when sounded together. $A$ gas column $37.5 \, cm$ long in a pipe closed at one end resonates in its fundamental mode with fork $A$,whereas a column of length $38.5 \, cm$ of the same gas in a similar pipe is required for resonance with fork $B$. The frequencies of these two tuning forks are:

A
$308 \, Hz, 300 \, Hz$
B
$208 \, Hz, 200 \, Hz$
C
$300 \, Hz, 400 \, Hz$
D
$350 \, Hz, 500 \, Hz$

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Waves and Sound

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Beats and Tuning fork

$41$ tuning forks are arranged in increasing order of frequency such that each produces $5 \text{ beats/second}$ with the next tuning fork. If the frequency of the last tuning fork is double that of the frequency of the first fork,then the frequency of the first and last fork is:

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Two tuning forks $F_1$ and $F_2$ produce $6$ beats per second when sounded together. The frequency of $F_1$ is $256 \ Hz$. When wax is applied to $F_2$,the number of beats per second remains $6$. What is the frequency of $F_2$ in $Hz$?

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$50$ tuning forks are arranged in increasing order of their frequencies such that each gives $4 \, \text{beats/sec}$ with its previous tuning fork. If the frequency of the last fork is the octave of the first, then the frequency of the first tuning fork is ... $Hz$.

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The frequency of tuning fork $A$ is $256\,Hz$. It produces $4\,beats/sec$ with tuning fork $B$. When wax is applied to tuning fork $B$,$6\,beats/sec$ are heard. By reducing a small amount of wax,$4\,beats/sec$ are heard. The frequency of $B$ is .... $Hz$.

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The frequencies of two sound sources are $256 \text{ Hz}$ and $260 \text{ Hz}$. At $t = 0$,the intensity of sound is maximum. Then the phase difference at the time $t = \frac{1}{16} \text{ s}$ will be

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$41$ tuning forks are arranged in increasing order of frequency such that each produces $5 \text{ beats/second}$ with the next tuning fork. If the frequency of the last tuning fork is double that of the frequency of the first fork,then the frequency of the first and last fork is:

Two tuning forks $F_1$ and $F_2$ produce $6$ beats per second when sounded together. The frequency of $F_1$ is $256 \ Hz$. When wax is applied to $F_2$,the number of beats per second remains $6$. What is the frequency of $F_2$ in $Hz$?

$50$ tuning forks are arranged in increasing order of their frequencies such that each gives $4 \, \text{beats/sec}$ with its previous tuning fork. If the frequency of the last fork is the octave of the first, then the frequency of the first tuning fork is ... $Hz$.

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