Two uniform strings A and B made of steel are made to vibrate under the same tension. If the first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B, the ratio of the length of the strings is (1) 1:2 (2) 1:3 (3) 1:4 (4) 1:5

Answer: (2) 1:3

  • Frequency of A = n1
  • Frequency of B = n2

∴ 2n1 = 3n2 or

\(\begin{array}{l}\frac{2}{2l_{1}}\sqrt{\frac{T}{m_{1}}} = \frac{3}{2l_{2}}\sqrt{\frac{T}{m_{2}}}\end{array} \)
.

Where m = mass/length

m = (cross sectional area × l × ρ)/l

m = cross sectional area × ρ

\(\begin{array}{l}\frac{l_{1}}{l_{2}}= \frac{2}{3}\sqrt{\frac{m_{2}}{m_{1}}}= \sqrt{\frac{a_{2}\rho}{a_{1}\rho}}\end{array} \)

Where

  • a = Cross-sectional area(a = πr2)
  • ρ = Density of the material

Since density is same, above equation becomes

\(\begin{array}{l}\frac{l_{1}}{l_{2}}= \frac{2}{3}\sqrt{\frac{r_{2}^{2}}{r_{1}^{2}}}\end{array} \)
.

\(\begin{array}{l}\frac{l_{1}}{l_{2}}= \frac{2}{3}\sqrt({\frac{1}{2})^{2}}\end{array} \)
.

\(\begin{array}{l}\frac{l_{1}}{l_{2}}= \frac{1}{3}\end{array} \)

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