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The velocity of waves in a string fixed at both ends is $2 \mathrm{~m} / \mathrm{s}$.
The string forms standing waves with nodes $5.0 \mathrm{~cm}$ apart. The frequency of vibration of the string in $\mathrm{Hz}$ is
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The string forms standing waves with nodes $5.0 \mathrm{~cm}$ apart. The frequency of vibration of the string in $\mathrm{Hz}$ is
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The correct answer is:
20
Here the distance between the two nodes is half of the wavelength
\(\frac{\lambda}{2}=5.0 \mathrm{~cm} \Rightarrow \lambda=10 \mathrm{~cm}\)
Hence \(\mathrm{n}=\frac{\mathrm{v}}{\lambda}=\frac{200}{10}=20 \mathrm{~Hz}\)
\(\frac{\lambda}{2}=5.0 \mathrm{~cm} \Rightarrow \lambda=10 \mathrm{~cm}\)
Hence \(\mathrm{n}=\frac{\mathrm{v}}{\lambda}=\frac{200}{10}=20 \mathrm{~Hz}\)
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