Question

In a sonometer experiment, the fundamental frequency is n. If the distance between the two bridges and tension both are doubled from the original value, the frequency will become
A. \[\sqrt{2}n\]
B. \[2n\]
C. \[\left( {1}/{2}\; \right)n\]
D. \[\left( {1}/{\sqrt{2}}\; \right)n\]

Answer 1

Hint: Recall what is frequency and tension and try to find out the relation between them. In the equation \[n=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}\], replace \[T\]and \[n\] with \[{{T}^{'}}=2T\] and \[{{l}^{'}}=2l\] respectively. Compare it with \[n\].

Complete step-by-step answer:
We are familiar with the terms like frequency, amplitude and wavelength. These are the parameters to represent a wave. Wavelength is the distance between two consecutive crests or two successive troughs.
Amplitude is the maximum displacement of the particle from the mean position. Frequency is the number of waves passing through a particular point in one-second or number of cycles in one second. Usually, frequency is denoted by ‘\[n\]’. The unit of frequency is \[{{s}^{-1}}\] or Hertz (Hz).
Tension is a restoring force that develops on a body when it is under the influence of some external forces. For example, if we try to stretch a string with our hand, then the tension will be generated in that spring to oppose the external force.
There is a device which helps to explain the relationship between frequency and tension. We use a sonometer to show the relation between these two. It consists of a wooden board on which a stretched copper or brass wire is placed whose one end is attached to a weight. It finds the frequency of a tuning, to find tension etc.
The working principle is that for a constant vibrating length and mass per unit length, the frequency of the stretched vibrating is proportional to the tension created in the string. Then the equation is given as, i.e. \[n=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}\]………………………..(1)
where m is the mass/unit length
Where, \[l\] is the distance between two bridges.
In the problem, the tension is doubled.i.e \[{{T}^{'}}=2T\] and
bridge length is also increased by \[{{l}^{'}}=2l\]. Substituting in equation (1)
\[{{n}^{'}}=\dfrac{1}{2(2l)}\sqrt{\dfrac{2T}{m}}\] or
\[{{n}^{'}}=\dfrac{\sqrt{2}}{2}\left( \dfrac{1}{2l}\sqrt{\dfrac{T}{m}} \right)\] ………………..(2)
Comparing equation(1) and (2). We get
\[{{n}^{'}}=\dfrac{\sqrt{2}}{2}n\] or simplifying
\[{{n}^{'}}=\dfrac{1}{\sqrt{2}}n\] or
\[{{n}^{'}}=\left( {1}/{\sqrt{2}}\; \right)n\], the new frequency when both the tension and bridge lengths are doubled.
Then option D is the answer.

Note: From the above problem what we can observe is that the frequency is reduced by \[\dfrac{1}{\sqrt{2}}\], i.e., 0.707. As the length increases, the frequency of the transverse generated decreases. But it increases with the tension. The tension created in spring is a scalar quantity.