What length of this wire should be plucked to cause it to vibrate with a frequency of 840 Hz?

Solution

Given,

Diameter of the wire (d) = 0.04 cm = 0.04 x 10^-2 m
Density of wire (ρ) = 8000 kgm^-3 
Tension in the wire (T) = 80 N
Frequency (f) = 840 Hz
Length of the wire (l) = ?

We know that the frequency of vibration (f) is given by

\(f=\frac1{2l}\sqrt{\frac Tm}=\frac1{2l}\sqrt{\frac T{\displaystyle\frac{\rho\pi d^2}4}}\)
Or, \(f=\frac1{ld}\sqrt{\frac T{\pi\rho}}\Rightarrow l=\frac1{fd}\sqrt{\frac T{\pi\rho}}\)
Or, \(l=\frac1{840\times0.04\times10^{-2}}\sqrt{\frac{80}{\pi\times8000}}=0.168\)

Hence, the length of the wire is 0.168 m.