Question

A sonometer wire, with a suspended mass of \[M = 1\,{\text{kg}}\], is in resonance with a given tuning fork. The apparatus is taken to the moon where the acceleration due to gravity is \[\dfrac{1}{6}\] that on earth. To obtain resonance on moon, the value of \[M\] should be
A. \[1\,{\text{kg}}\]
B. \[\sqrt 6 \,{\text{kg}}\]
C. \[6\,{\text{kg}}\]
D. \[36\,{\text{kg}}\]

Answer 1

Hint: Use the formula for the resonance frequency. Also substitute the force balanced by the tension in sonometer wire in the formula to determine the value of M.

Formula used:
The fundamental frequency \[f\] is given by
\[f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} \] …… (1)

Here, \[l\] is the length of the string, \[T\] is the tension on the string and \[m\] is the linear density of the string.


Complete step by step answer:
The suspended mass \[M\] on the sonometer wire is \[1\,{\text{kg}}\].
\[M = 1\,{\text{kg}}\]

The resonance tension \[T\] of the sonometer wire is balanced by the weight \[Mg\] of the suspended mass.
\[T = Mg\]

Determine the resonance frequency for the sonometer wire on the earth.

Substitute \[Mg\] for \[T\] in equation (1).
\[f = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{m}} \]

The length and linear density of the sonometer wire and the mass of the suspended mass remains the same on earth and the moon. But the resonance frequency changes on the moon.

In order to obtain the same resonance frequency on the moon, the suspended mass should be changed to \[M'\].

Rewrite the above equation for the resonance frequency \[f'\] on the moon.
\[f' = \dfrac{1}{{2l}}\sqrt {\dfrac{{M'g'}}{m}} \]

Here, \[g'\] is the acceleration due to gravity on the moon.

The acceleration due to gravity \[g'\] on the moon is \[\dfrac{1}{6}\] that on the earth.
\[g' = \dfrac{1}{6}g\]

We want to obtain the same resonance frequency on the earth and the moon.
\[f = f'\]

Substitute \[\dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{m}} \] for \[f\] and \[\dfrac{1}{{2l}}\sqrt {\dfrac{{M'g'}}{m}} \] for \[f'\] in the above equation.
\[\dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{m}} = \dfrac{1}{{2l}}\sqrt {\dfrac{{M'g'}}{m}} \]
\[ \Rightarrow \sqrt {Mg} = \sqrt {M'g'} \]
\[ \Rightarrow Mg = M'g'\]

Rearrange the above equation for \[M'\].
\[M' = \dfrac{{Mg}}{{g'}}\]

Substitute \[\dfrac{1}{6}g\] for \[g'\] and \[1\,{\text{kg}}\] for \[M\] in the above equation..
\[M' = \dfrac{{\left( {1\,{\text{kg}}} \right)g}}{{\dfrac{1}{6}g}}\]
\[ \Rightarrow M' = 6\,{\text{kg}}\]

Therefore, the mass \[M\] on the moon should be \[6\,{\text{kg}}\].

Hence, the correct option is C.

Note:The mass of the suspended mass remains the same on the earth and the moon. But in order to have the same resonance frequency on the earth and the moon, the mass \[M\] should be changed.