Question

Derive f=1/2l√t/m dimensionally t is the tension f is the frequency l is length and m is the mass per unit length

Answer 1

To derive:

$\boxed{f = \dfrac{1}{2l} \sqrt{ \dfrac{t}{m} } } \: \: \: \: \: \: \: \: ......(dimensional \: analysis)$

Derivation:

Let frequency be f be dependent on these quantities as follows :

$\therefore \: f \propto \: {l}^{x} \: {t}^{y} \: {m}^{z}$

Now , expressing them in terms of basic physical quantities :

$\implies\: {T}^{ - 1} \propto \: {L}^{x} \: \times {(ML{T}^{ - 2}) }^{y} \: \times {(M{L}^{ - 1} )}^{z}$

$\implies\: {T}^{ - 1} \propto \: {M}^{(y + z)} \times {L}^{(x + y - z)} \times {T}^{ - 2y}$

So, comparing both sides:

$1) \: - 2y = - 1 \\ 2) \: y + z = 0 \\ 3) \: x + y - z = 0$

So, after solving these equations , we get:

$1) \: x = - 1 \\ 2) \: y = \dfrac{1}{2} \\ 3) \: z = - \dfrac{1}{2}$

So, putting value of x , y , z :

$\therefore \: f \propto \: {l}^{ - 1} \: {t}^{ \frac{1}{2} } \: {m}^{ - \frac{1}{2} }$

$\implies\: f = \:k \: \bigg( {l}^{ - 1} \: {t}^{ \frac{1}{2} } \: {m}^{ - \frac{1}{2} } \bigg)$

$\implies\: f = \:k \: \bigg( \dfrac{1}{l} \sqrt{ \dfrac{t}{m} } \bigg)$

Hence , derived ✔️