A wire having a diameter of 0.90 mm is replaced by another wire of the same material but with a diameter of 0.93 mm. If the tension of the wire is as before what is the percent change in the frequency of fundamental note?

Solution

Given,

1^st diameter (d₁) = 0.9 mm = 0.9 x 10-3 m
2^nd diamter (d₂) = 0.93 mm = 0.93 x 10-3 m

The fundamental frequency of vibration of wire is given as

\(f=\frac1{ld}\sqrt{\frac T{\pi\rho}}\)

Similarly ,

\(f_1=\frac1{ld_1}\sqrt{\frac T{\pi\rho}}\)

And,

\(f_2=\frac1{ld_2}\sqrt{\frac T{\pi\rho}}\)

Percentage change in frequency =\(=\frac{f_1-f_2}2\times100\%\)

\(=\frac{{\displaystyle\frac1{ld_1}}\sqrt{\displaystyle\frac T{\pi\rho}}-{\displaystyle\frac1{ld_2}}\sqrt{\displaystyle\frac T{\pi\rho}}}{{\displaystyle\frac1{ld_1}}\sqrt{\displaystyle\frac T{\pi\rho}}}\)
\(=\frac{{\displaystyle\frac1{d_1}}-{\displaystyle\frac1{d_2}}}{\displaystyle\frac1{d_1}}\times100\%\)
\(=\frac{d_2-d_1}{d_1d_2}\times d_1\times100\%\)
\(=\frac{0.93-0.9}{0.93}\times100\%\)
= 3.2%

Thus, the required percentage change is 3.2%