A sound of frequency $480 \mathrm{~Hz}$ is emitted from the stringed instrument. The velocity of sound in…
A sound of frequency $480 \mathrm{~Hz}$ is emitted from the stringed instrument. The velocity of sound in air is $320 \mathrm{~m} / \mathrm{s}$. After completing 180 vibrations, the distance covered by a wave is
$60 \mathrm{~m}$
$90 \mathrm{~m}$
$120 \mathrm{~m}$
$180 \mathrm{~m}$
Solution
Given: $\mathrm{v}=320 \mathrm{~m} / \mathrm{s}, \mathrm{f}=480 \mathrm{~Hz}, \mathrm{~N}=180$
$\mathrm{v}=\mathrm{f} \lambda$
$\therefore \quad \lambda=\frac{\mathrm{v}}{\mathrm{f}}$
Substituting the values, we get
$\lambda=\frac{320}{480}=\frac{2}{3}$
$\therefore \quad$ The total distance covered after 180 vibrations is
$\begin{aligned}
& \mathrm{D}=\mathrm{N} \times \lambda \\
& \mathrm{D}=180 \times \frac{2}{3} \\
& \mathrm{D}=120 \mathrm{~m}
\end{aligned}$