Two sounding sources send waves at certain temperature in air of wavelength $50 \mathrm{~cm}$ and $50.5…
Two sounding sources send waves at certain temperature in air of wavelength $50 \mathrm{~cm}$ and $50.5 \mathrm{~cm}$ respectively. The frequency of sources differ by $6 \mathrm{~Hz}$. The velocity of sound in air at same temperature is
$300 \mathrm{~m} / \mathrm{s}$
$303 \mathrm{~m} / \mathrm{s}$
$313 \mathrm{~m} / \mathrm{s}$
$330 \mathrm{~m} / \mathrm{s}$
Solution
$\mathrm{v}=\mathrm{n} \lambda$
Since, both the sound sources are at same temperature, velocity of sound in both cases would be the same.
$\begin{aligned}
& \therefore \quad \mathrm{v}=\left(50 \mathrm{n}_{\mathrm{l}}\right) \mathrm{cm} / \mathrm{s} .....(i) \\
& \mathrm{v}=\left(50.5 \mathrm{n}_2\right) \mathrm{cm} / \mathrm{s} ....(ii)\\
& \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{50.5}{50} ....[From (i) and (ii)]\\
& \therefore \quad \frac{\mathrm{n}_1-\mathrm{n}_2}{\mathrm{n}_2}=\frac{50.5-50}{50} \\
&
\end{aligned}$
$\begin{array}{ll}\therefore & \frac{6}{\mathrm{n}_2}=\frac{0.50}{50}=\frac{1}{100} \quad \ldots .\left(\because \mathrm{n}_1-\mathrm{n}_2=6 \mathrm{~Hz}\right) \\ \therefore & \mathrm{n}_2=600 \mathrm{~Hz} \\ \therefore & \mathrm{v}=\frac{50.5 \times 600}{100} \mathrm{~m} / \mathrm{s} \ldots[\text { From (ii) }] \\ \therefore & \mathrm{v}=303 \mathrm{~m} / \mathrm{s}\end{array}$