The suspended mass \(M\) on the sonometer wire is \(1 \mathrm{~kg}\).
\(M=1 \mathrm{~kg}\)
The resonance tension \(T\) of the sonometer wire is balanced by the weight \(M g\) of the suspended mass.
\(T=M g\)
Substitute \(M g\) for \(T\) in equation (1).
\(f=\frac{1}{2 l} \sqrt{\frac{M g}{m}}\)
Rewrite the above equation for the resonance frequency \(f^{\prime}\) on the moon.
\(f^{\prime}=\frac{1}{2 l} \sqrt{\frac{M^{\prime} g^{\prime}}{m}}\)
Here, \(g^{\prime}\) is the acceleration due to gravity on the moon.
The acceleration due to gravity \(g^{\prime}\) on the moon is \(\frac{1}{6}\) that on the earth.
\(g^{\prime}=\frac{1}{6} g\)
We want to obtain the same resonance frequency on the earth and the moon.
\(f=f^{\prime}\)
Substitute \(\frac{1}{2 l} \sqrt{\frac{M g}{m}}\) for \(f\) and \(\frac{1}{2 l} \sqrt{\frac{M^{\prime} g^{\prime}}{m}}\) for \(f^{\prime}\) in the above equation.
\(\frac{1}{2 l} \sqrt{\frac{M g}{m}}=\frac{1}{2 l} \sqrt{\frac{M^{\prime} g^{\prime}}{m}}\)
\(\Rightarrow \sqrt{M g}=\sqrt{M^{\prime} g^{\prime}}\)
\(\Rightarrow M g=M^{\prime} g^{\prime}\)
Rearrange the above equation for \(M^{\prime}\).
\(M^{\prime}=\frac{M g}{g^{\prime}}\)
Substitute \(\frac{1}{6} g\) for \(g^{\prime}\) and \(1 \mathrm{~kg}\) for \(M\) in the above equation.
\(M^{\prime}=\frac{(1 \mathrm{~kg}) g}{\frac{1}{6} g}\)
\(\Rightarrow M^{\prime}=6 \mathrm{~kg}\)
Therefore, the mass \(M\) on the moon should be \(6 \mathrm{~kg}\).