The total length of a sonometer wire between fixed ends is \(110\, cm\). Two bridges are placed to divide the length of wire in ratio \(6 : 3 : 2\). The tension in the wire is \(400\, N\) and the mass per unit length is \(0.01\, kg/m\) . What is the minimum common frequency with Which three parts can vibrate ........... \(Hz\) ?

Length of wire is in ratio,\(6: 3: 2\) i.e \(60 \mathrm{cm}\), \(30 \mathrm{cm}, 20 \mathrm{cm}\)

Tension in the wire, \(T=400 \mathrm{N}\)

Mass per unit length, \(\mathrm{m}=0.01 \mathrm{kg}\)

Minimum common frequency \(=?\)

As we know,

Frequency, \(v=\frac{1}{21} \sqrt{\frac{T}{m}}=\frac{1000}{11}-H z\)

Similarly, \(v_{1}=\frac{1000}{6} \mathrm{Hz}\)

\(v_{2}=\frac{1000}{3} \mathrm{Hz}\)

\(v_{3}=\frac{1000}{2} \mathrm{Hz}\)

Hence common frequency \(=1000 \mathrm{Hz}\)