A string A has twice the length, twice the diameter, twice the tension and twice the density of another string B. The overtone of A which will, have the same fundamental frequency as that of B is

Step-by-step solution

Fundamental frequency of string B is, aligned n _ B & = 1 2 l_ B ~T _ B _ B . where, = mass per unit length. m _ B & = r ^2 l _ B & = r ^2 l l = r ^2 _ A & =2 (2 r )^2=8 _ B _ A & = 8 f r ^2 2 l =4 _ B aligned Fundamental frequency of string A is, n _ A = 1 2 (2 l_ B ) 2 ~T _ B 8 _ B = 1 4 ( 1 2 l_ B ~T _ B _ B )= 1 4 n _ B Frequency of p ^ th overtone =(p+1) n Here, (p+1)=4 p=3 3^ rd Overtone of string A will have same frequency as fundamental frequency of string B .