First slide

Standing waves

Question

Two uniform strings, A and B made of steel are made to vibrate under the same tension. Radius of A is two times the radius of B . If the first overtone of A is equal to the second overtone of B, the ratio of lengths of the string is

Easy

A

1 :2
B

1 :3
C

1:4
D

1:5

Solution

Given that first overtone of string A is equal to second overtone of string 8. Thus the second harmonic of A is equal to third harmonic of B, i.e.,
$or 2 [\frac{1}{L_{A} r_{A}} \sqrt{(\frac{T}{πρ})}] = 3 [\frac{1}{L_{B} r_{B}} \sqrt{(\frac{T}{πρ})}] or 2 [\frac{1}{L_{A} r_{A}} \sqrt{(\frac{T}{ρ})}] = 3 [\frac{1}{L_{B} r_{B}} \sqrt{(\frac{T}{ρ})}] or \frac{2}{L_{A} r_{A}} = \frac{3}{L_{B} r_{B}} or \frac{L_{A}}{L_{B}} = \frac{2 r_{B}}{3 r_{A}} = \frac{2 (r_{B})}{3 (2 r_{B})} or \frac{L_{A}}{L_{B}} = \frac{1}{3} or L_{A} : L_{B} = 1 : 3$

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