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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.
Solution:
1241 Upvotes
Verified Answer
The correct answer is:
5
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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