JEE Advanced · Physics · 16. Waves & Sound
A \(20 \mathrm{~cm}\) long string, having a mass of \(1.0 \mathrm{~g}\), is fixed at both the ends. The tension in the string is \(0.5 \mathrm{~N}\). The string is set into vibration using an external vibrator of frequency 100 \(\mathrm{Hz}\). Find the separation (in \(\mathrm{cm}\) ) between the successive nodes on the string.
- A 1
- B 2
- C 3
- D 5
Answer & Solution
Correct Answer
(D) 5
Step-by-step Solution
Detailed explanation
Distance between the successive nodes,
\(
\begin{aligned}
d & =\frac{\lambda}{2}=\frac{v}{2 f} \\
& =\frac{\sqrt{T / \mu}}{2 f}
\end{aligned}
\)
Substituting the values we get
\(
d=5 \mathrm{~cm}
\)
\(
\begin{aligned}
d & =\frac{\lambda}{2}=\frac{v}{2 f} \\
& =\frac{\sqrt{T / \mu}}{2 f}
\end{aligned}
\)
Substituting the values we get
\(
d=5 \mathrm{~cm}
\)
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