Solution
Given,
Length of the cord (l) = 1.5 m
Mass per unit length (μ) = 1.2 g/m = \(\frac{1.2}{1000}\) kg/m
Tension on cord (T) = 12 N
- Fundamental frequency (f) = ?
We have, \(f=\frac1{2l}\sqrt{\frac T\mu}=\frac1{2\times1.5}\sqrt{\frac{12\times1000}{1.2}}=33.33\) Hz
Again, for n = 3 mode, let f0 be the fundamental frequency. Here,
f0 = 0.5 KHz = 0.50 x 1000 Hz ,
T = ?
Now,
\(f_0=\frac1{2l}\sqrt{\frac T\mu}\)
Or, \(\frac{500}3=\frac1{2\times1.5}\sqrt{\frac{T\times1000}{1.2}}\)
Or, \(T=\frac{1.2}{1000}\times\left(\frac{500\times2\times1.2}3\right)^2=\frac{1.2}{1000}\times\left(500\right)^2=300\) N
∴ The tension required on the string when n = 3 with frequency of 0.5 kHz is 300 N