Question 236: Waves

Question

Fundamental frequency of sonometer wire is ' $n$ '. If the tension and length are increased 3 times and diameter is increased twice, the new frequency will be

Accepted Answer

The frequency $ n $ of a sonometer wire is given by the formula:
$ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $

where $ L $ is the length of the wire, $ T $ is the tension in the wire, and $ \mu $ is the mass per unit length.
The mass per unit length, $ \mu $, can also be written as:

$ \mu = 	ext{density} 	imes 	ext{area} = ho \cdot \frac{\pi d^2}{4} $

This shows that $ \mu $ is proportional to $ d^2 $ (the square of the diameter).
So, the frequency can be related to different variables as:

$ n \propto \frac{1}{L} \sqrt{\frac{T}{d^2}} $
Now let's apply the changes in the question:
Tension $ T $ becomes 3 times bigger.
Length $ L $ becomes 3 times bigger.
Diameter $ d $ becomes 2 times bigger.
Let the new frequency be $ n' $. Then:

$ \frac{n'}{n} = \frac{L}{L'} 	imes \sqrt{\frac{T'}{T}} 	imes \frac{d}{d'} $
Substitute the new values:
$ L' = 3L $, so $ \frac{L}{L'} = \frac{1}{3} $
$ T' = 3T $, so $ \sqrt{\frac{T'}{T}} = \sqrt{3} $
$ d' = 2d $, so $ \frac{d}{d'} = \frac{1}{2} $
Put these values into the formula:

$ \frac{n'}{n} = \frac{1}{3} 	imes \sqrt{3} 	imes \frac{1}{2} $
This simplifies to:

$ \frac{1}{3} 	imes \frac{1}{2} 	imes \sqrt{3} = \frac{1}{6} \sqrt{3} $

But according to the original, we use $ \frac{1}{3} 	imes \sqrt{\frac{3}{2^2}} = \frac{1}{3} 	imes \sqrt{\frac{3}{4}} = \frac{1}{3} 	imes \frac{\sqrt{3}}{2} = \frac{1}{2\sqrt{3}} $.
So, the new frequency is:

$ n' = \frac{n}{2\sqrt{3}} $

Answer

$2 n$

Answer

$\frac{\sqrt{3}}{2} n$

Answer

$\frac{n}{2 \sqrt{3}}$

Answer

$\sqrt{3} n$

Waves

Fundamental frequency of sonometer wire is ' n '. If the tension and length are increased 3 times and diameter is increased twice, the new frequency will be

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