Question 406: Waves

Question

A hollow pipe of length $$0.8 \mathrm{~m}$$ is closed at one end. At its open end, a $$0.5 \mathrm{~m}$$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of pipe. If the tension in the string is $$50 \mathrm{~N}$$ and speed of sound in air is $$320 \mathrm{~m} / \mathrm{s}$$, the mass of the string is

Accepted Answer

The fundamental frequency of the closed pipe $$(n)=\frac{V}{4 L}$$
$$\mathrm{n}=\frac{320}{4 	imes 0.8}=\frac{320}{3.2}=100 \mathrm{~Hz}$$
For the vibrating wire, fundamental frequency (n) is
$$\mathrm{n}^{\prime}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$$
$$	herefore$$ For the second harmonic, frequency $$=2 n^{\prime}$$
$$=\frac{2}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{1}{\mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$$
It is given that $$2 \mathrm{n}^{\prime}=\mathrm{n}=100$$ (Resonance)
$$\begin{aligned}
& 	herefore 100=\frac{1}{\mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{1}{0.5} \sqrt{\frac{50}{\mathrm{~m}}} \
& 	herefore 100 	imes 0.5=\sqrt{\frac{50}{\mathrm{~m}}} \quad 	ext { on squaring }
\end{aligned}$$
$$\begin{aligned}
& 	herefore(50)^2=\frac{50}{\mathrm{~m}} \
& 	herefore 50 \mathrm{~m}=1 \
& 	herefore \mathrm{m}=\frac{1}{50} \mathrm{~kg}=\frac{1}{50} 	imes 1000 \mathrm{~g}=20 	ext { gram }
\end{aligned}$$

Answer

20 g

Answer

10 g

Answer

40 g

Answer

5 g

Waves

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