Question

In a sonometer experiment, resonance occurs when the length of the wire between bridges is $20\,cm$ and tuning fork of frequency $300\,Hz$ excites the vibrations. If a tuning fork of frequency $400\,Hz$ is used, thin the resonant length of the wire will be
A. 10 cm
B. 15 cm
C. 20 cm
D. 25 cm

Answer 1

Hint: The sonometer is a wooden box with a long string that is used to suspend the load. The frequency of vibration of the string depends on the length, tension in the string and the linear density of the string as well. Hence knowing the exact relation between the frequency of the sonometer wire and the linear density of the wire will enable us to select the correct alternative.

Formula used:
$f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $

Complete step by step answer:
So as from the formula of sonometer the frequency is inversely proportional to the length.
It is given that frequencies ${f_{1{\text{ }}}}and{\text{ }}{f_2}$ are 300 Hz and 400 Hz respectively and length ${l_1}$ is $20{\text{ cm}}$. And we have to find length ${l_2}$.Therefore, taking the ratio of frequencies 1 and 2 we get,
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{{l_2}}}{{{l_1}}}$
Therefore,
$ \Rightarrow {l_2} = \dfrac{{300}}{{400}} \times 20$
$\therefore {l_2} = 15{\text{ cm}}$

Therefore, the correct answer is option B.

Additional information: Mass per unit length of a wire is an intrinsic property. Hence different materials will have m different. When the frequency of the applied source is equal to the natural frequency of the sonometer wire for given conditions, the wire vibrates with maximum frequency due to resonance. Hence when conducting the experiment for a particular source frequency, the amplitude of vibration is maximum.

Note: Frequency is defined as the number of oscillations or occurrences per unit time. The unit of frequency is Hertz. Frequency is also defined as the reciprocal of time period. From the above calculation you can see that by increasing the length and diameter of the cross-section of wire the frequency decreases.