Question

Prove the following using dimentional analysis

v = 1 / 2l + sqrt{t/n} 

v = frequency T= tension force n = mass per length.

Answer 1

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Fundamental resonance frequency of a transverse wave on a thin rod or rope that has a uniform mass per unit length n is given by the following formula.

T is the force of tension in the rope or string or thin rod of length L. The string is fixed firmly at both ends or open at both ends.

$\nu\ =\ fundamental\ resonance\ frequency\\ \\ = \frac{1}{2L} \sqrt{ \frac{T}{\mu}} \\ \\ \left[\begin{array}{ccc} T^{-1} \end{array}\right]$

$= \left[\begin{array}{ccc} L^{-1} \end{array}\right] \left[\begin{array}{ccc} Force^{\frac{1}{2}} \end{array}\right] \left[\begin{array}{ccc} (M\ L^{-1})^{-\frac{1}{2}} \end{array}\right]$

[tex] = \left[\begin{array}{ccc} L^{-1} \end{array}\right]\ \left[\begin{array}{ccc} (M\ L\ T^{-2})^{\frac{1}{2}} \end{array}\right]\ \left[\begin{array}{ccc} (M\ L^{-1})^{-\frac{1}{2}} \end{array}\right] \\ [/tex]

[tex] = \left[\begin{array}{ccc} (M^{\frac{1}{2}-\frac{1}{2}}\ L^{-1+\frac{1}{2}+\frac{1}{2}}\ T^{2*\frac{1}{2}} \end{array}\right] \\ \\ = \left[\begin{array}{ccc} M^0\ L^0\ T^{-1} \end{array}\right] \\ [/tex]

So the dimensions on both sides match.