JEE Mains · Physics · STD 11 - 14. waves and sound
A string of length \(1\, m\) and mass \(5\, g\) is fixed at both ends. The tension in the string is \(8.0\, N\). The string is set into vibration using an external vibrator of frequency \(100\, Hz\). The separation between successive nodes on the string is close to .... \(cm\)
- A \(10\)
- B \(33.3\)
- C \(16.6\)
- D \(20.0\)
Answer & Solution
Correct Answer
(D) \(20.0\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}=\frac{\mathrm{n}}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\) On solving, \(n=5\) \(5\) loops are formed in \(1 \mathrm{m}\) Separation between successive nodes \(=\frac{1}{5} \mathrm{m}=20 \mathrm{cm}\)
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