Question

A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of \[9\;kg\] is suspended from the wire. When this mass is replaced by a mass \[M\] , the wire resonates with the same tuning fork forming three antinodes for the same position of the bridges. The value of \[M\] is
A. \[25\;kg\]
B. \[5\;kg\]
C. \[12.5\;kg\]
D. \[\dfrac{1}{{25}}\;kg\]

Answer 1

Hint: The sonometer is a wooden box with a long string that can be used to hang the load. The frequency of vibration of the string is determined by the length, tension in the string, and the linear density of the string too. Therefore if we know the exact relation between the frequency of the sonometer wire and the linear density of the wire will allow us to choose the exact substitute.

Complete step by step solution:
A sonometer is a device that is used to check the relationship between the wavelength and the frequency of the sound waves. The sonometer mechanism is based on the principle of transverse vibrations of a string when pulled.
Let us consider the length of the sonometer wire as the tension in the string is , the mass per unit length of the wire is$\mu $ and the number of poles is given by $P$ Then the frequency \[\left( f \right)\]of vibration of the wire is given by,
$f = \dfrac{P}{{2L}}\sqrt {\dfrac{T}{\mu }} $
The number of poles is equal to the number of Antinodes hence $P = 5$ and tension $T = 9g$,
hence ${f_1} = \dfrac{5}{{2L}}\sqrt {\dfrac{{9g}}{\mu }} $
and for $P = 3$
we get ${f_2} = \dfrac{3}{{2L}}\sqrt {\dfrac{{Mg}}{\mu }} $
for tension $T = Mg$
Equating both \[{f_1}\] and \[{f_2}\] we get $\dfrac{5}{{2L}}\sqrt {\dfrac{{9g}}{\mu }} = \dfrac{3}{{2L}}\sqrt {\dfrac{{Mg}}{\mu }} $
which gives $M = 25kg$

Note:
The mass per unit length of a wire is an inherent property. Hence different elements will have different $\mu $. If the frequency of the functional source is equal to the natural frequency of the sonometer wire for specified situations, the wire vibrates with maximum frequency because of resonance. Therefore when developing the experiment for a certain source frequency, the amplitude of vibration is maximum.