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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 $24 \mathrm{~cm}$ and 25 $\mathrm{cm}$ respectively. Calculate the frequencies of forks.
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Verified Answer
The correct answer is:
$150 \mathrm{~Hz}, 144 \mathrm{~Hz}$
Let the frequency of the first fork be $\mathrm{f}_{1}$ and that of second be $\mathrm{f}_{2}$
We then have, $\mathrm{f}_{1}=\frac{\mathrm{v}}{4 \times 24}$ and $\mathrm{f}_{2}=\frac{\mathrm{v}}{4 \times 25}$
We also see that $\mathrm{f}_{1}>\mathrm{f}_{2}$
$$
\therefore \mathrm{f}_{1}-\mathrm{f}_{2}=6
$$
and $\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{24}{25}$
Solving (i) and (ii), we get $\mathrm{f}_{1}=150 \mathrm{~Hz}$ and $\quad \mathrm{f}_{2}=144 \mathrm{~Hz}$
We then have, $\mathrm{f}_{1}=\frac{\mathrm{v}}{4 \times 24}$ and $\mathrm{f}_{2}=\frac{\mathrm{v}}{4 \times 25}$
We also see that $\mathrm{f}_{1}>\mathrm{f}_{2}$
$$
\therefore \mathrm{f}_{1}-\mathrm{f}_{2}=6
$$
and $\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{24}{25}$
Solving (i) and (ii), we get $\mathrm{f}_{1}=150 \mathrm{~Hz}$ and $\quad \mathrm{f}_{2}=144 \mathrm{~Hz}$
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