When the string of a sonometer of length $L$ between the bridges vibrates in the second overtone, the…
When the string of a sonometer of length $L$ between the bridges vibrates in the second overtone, the amplitude of vibration is maximum at
$\left(\frac{L}{2}\right)$
$\left(\frac{L}{4}\right)$ and $\left(\frac{3L}{4}\right)$
$\frac{L}{6}$ and $\frac{3L}{6}$
$\frac{L}{8}, \frac{3L}{8}, \frac{5L}{6}$
Solution
Second overtone means three loops. The diagram depicts a standing wave between two fixed ends showing three antinodes labeled $A_1$, $A_2$, and $A_3$.
$\therefore 3 \frac{\lambda}{2} = L\text{ or }\lambda = \frac{2L}{3}$
Antinodes will be obtained at $\frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}\text{ or }\frac{L}{6}, \frac{3L}{6}\text{ and }\frac{5L}{6}$