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  • If the tension and diameter of a sonometer wire of fundamental frequency n are doubled, and the density is halved, then its fundamental frequency will become:<stron

If the tension and diameter of a sonometer wire of fundamental frequency n are doubled, and the density is halved, then its fundamental frequency will become:

Option: 1

\frac{n}{4}


Option: 2

\sqrt{2}n


Option: 3

n


Option: 4

\frac {n}{\sqrt{2}}


Answers (1)

best_answer

Mass per unit length of a wire, 

m = \pi r^2 \rho

\therefore Fundamental frequency,

n = \frac{1}{2l} \sqrt{\frac{T}{m}} = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}

After changing the tension and density of the sonometer wire, the frequency is:

n{}'= \frac{1}{2l (2r)} \sqrt{\frac{2T}{\pi (\frac{\rho}{2})}} = \frac{1}{4lr} \sqrt{\frac{4T}{\pi \rho}} = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}    = n

If the diameter is doubled, the radius is also doubled. Fundamental frequency remains the same.

Posted by

Ajit Kumar Dubey

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