Question

derive the relation f=1/2l√T/m​

Answer 1

Given:

f=1/2l√T/m​

$\text { L.H.S. }=\mathrm{v}=\left[\mathrm{T}^{-1}\right]=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}\right]$

$\text { R.H.S. }=1 / 2 I=\sqrt{T} / \mathrm{m}$

Tension in force

$\text { R.H.S. }=\frac{1}{L} \sqrt{\frac{M L T^{-2}}{M L^{-1}}}$

$\left.=\frac{1}{\mathrm{~L}} \sqrt{\mathrm{L}^{2} \mathrm{~T}^{-2}}=\frac{1}{\mathrm{~L}} \mid \mathrm{LT}^{-1}\right]$

$=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1} \mid\right.$

∴ L.H.S=R.H.S

Hence the formula is right

Answer 2

$v = 1/2l √T/m$

On both sides, write the measurements; we now have

$L.H.S. = v =[T-1] = [M0L0T-1]$

$R.H.S. = 1/2l = √T/m$

(Because 1/2 doesn't have any dimensions)

Tension is defined as force per unit length, while m is defined as mass per unit length.

$\text { R.H.S. } \begin{aligned}&=\frac{1}{L} \sqrt{\frac{M L T^{-2}}{M L^{-1}}} \\&\left.=\frac{1}{L} \sqrt{L^{2} T^{-2}}=\frac{1}{L} \mid \mathrm{LT}^{-1}\right] \\&=\left[M^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1} \mid\right.\end{aligned}$

Dimensionally, L.H.S. = R.H.S. As a result, the formulas are correct.