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Question

(A) In the second overtone of a string fixed at both ends,the string vibrates in $3$ loops (segments).For a string of length $L$ vibrating in $n$ loop

Vedclass · Accepted Answer

$\frac{L}{6}, \frac{L}{2}, \frac{5L}{6}$

Vedclass · Answer

(A) In the second overtone of a string fixed at both ends,the string vibrates in $3$ loops (segments).For a string of length $L$ vibrating in $n$ loops,the positions of the antinodes (where amplitude is maximum) are given by $x = \frac{(2k-1)L}{2n}$,where $k = 1, 2, ..., n$.Here,$n = 3$ (second overtone corresponds to the $3^{rd}$ harmonic).For $k = 1$: $x_1 = \frac{(2(1)-1)L}{2(3)} = \frac{L}{6}$.For $k = 2$: $x_2 = \frac{(2(2)-1)L}{2(3)} = \frac{3L}{6} = \frac{L}{2}$.For $k = 3$: $x_3 = \frac{(2(3)-1)L}{2(3)} = \frac{5L}{6}$.Thus,the amplitude is maximum at $\frac{L}{6}, \frac{L}{2}, 	ext{ and } \frac{5L}{6}$.

In a sonometer experiment,a string of length $L$ under tension vibrates in its second overtone between two bridges. The amplitude of vibration is maximum at:

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