A string fixed at both the ends forms standing wave with node separation of $5 \mathrm{~cm}$. If the…
A string fixed at both the ends forms standing wave with node separation of $5 \mathrm{~cm}$. If the velocity of the wave on the string is $2 \mathrm{~m} / \mathrm{s}$, then the frequency of vibration of the string is
$0.2 \mathrm{~Hz}$
$10 \mathrm{~Hz}$
$20 \mathrm{~Hz}$
$40 \mathrm{~Hz}$
Solution
Separation between consecutive nodes, $\frac{\lambda}{2}=5 \mathrm{~cm}$
$\therefore \quad \lambda=10 \mathrm{~cm}=0.1 \mathrm{~m}$
The frequency of vibration is given as:
$\mathrm{n}=\frac{\mathrm{v}}{\lambda}=\frac{2}{0.1}=20 \mathrm{~Hz}$
.