Question 379: Waves

Question

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

Accepted Answer

Given: $$\mathrm{v}=320 \mathrm{~m} / \mathrm{s}, \mathrm{f}=480 \mathrm{~Hz}, \mathrm{~N}=180$$
$$\begin{aligned}
\mathrm{v} & =\mathrm{f} \lambda \
	herefore \quad \lambda & =\frac{\mathrm{v}}{\mathrm{f}}
\end{aligned}$$
Substituting the values, we get
$$\lambda=\frac{320}{480}=\frac{2}{3}$$
$$	herefore \quad$$ The total distance covered after 180 vibrations is
$$\begin{aligned}
& D=N 	imes \lambda \
& D=180 	imes \frac{2}{3} \
& D=120 \mathrm{~m}
\end{aligned}$$

Answer

60 m

Answer

90 m

Answer

120 m

Answer

180 m

Waves

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

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