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A resonance tube completely filled with water has a small hole at the bottom.
Length of the tube is $0.8 \mathrm{~m}$. A vibrating tuning fork of frequency $500 \mathrm{~Hz}$ is held
near the open end of tube. Water is slowly removed from the bottom. The
maximum number of resonances heard will be (Neglect end correction. Speed of
sound in air $=340 \mathrm{~m} / \mathrm{s}$ )
Options:
Length of the tube is $0.8 \mathrm{~m}$. A vibrating tuning fork of frequency $500 \mathrm{~Hz}$ is held
near the open end of tube. Water is slowly removed from the bottom. The
maximum number of resonances heard will be (Neglect end correction. Speed of
sound in air $=340 \mathrm{~m} / \mathrm{s}$ )
Solution:
1244 Upvotes
Verified Answer
The correct answer is:
$2$
$$
\begin{array}{l}
\ell=0.8 \mathrm{~m} \quad \mathrm{f}=500 \mathrm{~Hz} \quad \mathrm{v}=340 \mathrm{~m} / \mathrm{s} \\
\mathrm{v}=\mathrm{f} \lambda \quad \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{340}{500}=\frac{34}{50}=\frac{34}{50}=\frac{17}{25} \mathrm{~m} \\
\frac{\lambda}{4}=\frac{17}{100}=0.17 \mathrm{~m} \\
\frac{3 \lambda}{4}=0.51 \mathrm{~m} \\
\frac{5 \lambda}{4}=0.85 \mathrm{~m}
\end{array}
$$
So, only 2 resonances will be heard.
\begin{array}{l}
\ell=0.8 \mathrm{~m} \quad \mathrm{f}=500 \mathrm{~Hz} \quad \mathrm{v}=340 \mathrm{~m} / \mathrm{s} \\
\mathrm{v}=\mathrm{f} \lambda \quad \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{340}{500}=\frac{34}{50}=\frac{34}{50}=\frac{17}{25} \mathrm{~m} \\
\frac{\lambda}{4}=\frac{17}{100}=0.17 \mathrm{~m} \\
\frac{3 \lambda}{4}=0.51 \mathrm{~m} \\
\frac{5 \lambda}{4}=0.85 \mathrm{~m}
\end{array}
$$
So, only 2 resonances will be heard.
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