Question 372: Waves

Question

A tuning fork of frequency $$220 \mathrm{~Hz}$$ produces sound waves of wavelength $$1.5 \mathrm{~m}$$ in air at N.T.P. The increase in wavelength when the temperature of air is $$27^{\circ} \mathrm{C}$$ is nearly $$\left(\sqrt{\frac{300}{273}}=1.05\right)$$

Accepted Answer

$$\begin{aligned}
& v_0=f \lambda_0=220 	imes 1.5 \
& v_0=330 \mathrm{~m} / \mathrm{s}
\end{aligned}$$
We know,
$$\begin{aligned}
& \frac{\mathrm{v}}{\mathrm{v}_0}=\sqrt{\frac{\mathrm{T}}{\mathrm{T}_0}} \
& \Rightarrow \mathrm{v}=330 \sqrt{\frac{300}{273}}=330 	imes 1.05 \
& \mathrm{v}=346.1 \mathrm{~m} / \mathrm{s} \
& 	herefore \quad \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{346.1}{220} \
& \lambda=1.57 \mathrm{~m}
\end{aligned}$$
$$	herefore \quad$$ The increase in wavelength is:
$$\begin{aligned}
& \Delta \lambda=\lambda-\lambda_0=1.57-1.5 \
& \Delta \lambda=0.07 \mathrm{~m}
\end{aligned}$$

Answer

0.06 m

Answer

0.10 m

Answer

0.09 m

Answer

0.07 m

Waves

A tuning fork of frequency $220 \mathrm{~Hz} produces sound waves of wavelength 1.5 \mathrm{~m} in air at N.T.P. The increase in wavelength when the temperature of air is 27^{\circ} \mathrm{C} is nearly \left(\sqrt{\frac{300}{273}}=1.05\right)

Choose the correct answer: