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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 : \(\frac{1}{2}\sqrt {\frac{T}{{ml}}} \)


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


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 \(\frac{1}{2}\sqrt {\frac{T}{{ml}}} \)

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