# A uniform string of length l and mass m is tied at both ends by the tension T, its frequency will be -

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A uniform string of length l and mass m is tied at both ends by the tension T, its frequency will be -
1. $\frac{1}{{2l}}\sqrt {\frac{T}{m}}$
2. $\frac{1}{2}\sqrt {\frac{T}{m}}$
3. $2l\sqrt {\frac{T}{m}}$
4. $\frac{1}{2}\sqrt {\frac{T}{{ml}}}$

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Correct Answer - Option 4 : $\frac{1}{2}\sqrt {\frac{T}{{ml}}}$

Concept:

The frequency of oscillation of a string stretched under tension T is given by

$f =\frac{1}{2l}\sqrt {\frac{T}{μ}}{}$

Where f = frequency of oscillation, n = number of loops, l = Length of string, T =Tension, and μ= linear density

Calculation:

Given, mass of string m

length of string l

linear mass density

$\mu = \frac{m}{l}$

$\implies f =\frac{1}{2l}\sqrt{\frac{T}{(m/l)}}{}$

$\implies f =\frac{1}{2l}\sqrt{\frac{Tl}{(m)}}{}$

$\implies f =\frac{1}{2}\sqrt{\frac{Tl}{(ml^2)}}{}$

Writing l = √ (l)2

$\implies f =\frac{1}{2}\sqrt{\frac{T}{(ml)}}{}$

So, the correct option is $\frac{1}{2}\sqrt {\frac{T}{{ml}}}$