A $20 \mathrm{~cm}$ long string, having a mass of $1.0 \mathrm{~g}$, is fixed at both the ends. The tension…
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
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}
$
!