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Two tuning forks $A$ and $B$ produce $8\,beats/s$ when sounded together. $A$ gas column $37.5\,cm$ long in a pipe closed at one end resonate to 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
$308\,Hz, 300\,Hz$
$208\,Hz, 200\,Hz$
$300\,Hz, 400\,Hz$
$350\,Hz, 500\,Hz$
Solution
Tunning fork $A$
$\frac{\lambda_{1}}{4}=37.5$
$\mathrm{n}_{1}=\frac{\mathrm{v}}{\lambda_{1}}=\frac{\mathrm{v}}{4 \times 37.5}$
Tunning fork $B$
$\frac{\lambda_{2}}{4}=38.5$
$\therefore \quad n_{2}=\frac{v}{\lambda_{2}}=\frac{v}{4 \times 38.5}$
$\mathrm{n}_{1}-\mathrm{n}_{2}=8=\frac{\mathrm{v}}{4 \times 37.5}-\frac{\mathrm{v}}{4 \times 38.5}=8$
$\mathrm{v}=8 \times 4 \times 37.5 \times 38.5$
$\mathrm{n}_{1}=308 \mathrm{\,Hz}$ and $\mathrm{n}_{2}=308-8=300 \mathrm{\,Hz}$
