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