Question

the velocity of transverse waves along a string may depend upon the length L of the string ,tension F in the string and mass per unit length M of the string. Derive a possible formula for the velocity dimensionally.
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Answer 1

$\textsf{\underline {\Large{Deducing Relations}}}$ :

Given, Velocity $\mathsf{(\:m{s} ^{-1}\:)}$ depends upon length L, tension F and mass M.

Let $\mathsf{v\:{\propto{\:{l} ^{a} \:{F} ^{b} \:{M{L} ^{-1}} ^{c}}}}$

Now, $\boxed{\mathsf{v\:=\:k\:{l} ^{a} \:{F} ^{b} \:{M{L} ^{-1}} ^{c}}}$ ↪️( i )

Here, K is a constant which is dimensionless.

Dimension of v = $\mathsf{[L{T} ^{-1}]}$

Dimension of l = $\mathsf{[L]}$

Dimension of F = $\mathsf{[ML{T} ^{-2}]}$

Dimension of M per unit length = $\mathsf{[M{L} ^{-1}]}$

Substituting these dimensions in equation ( i ),

$\mathsf{v\:=\:{l} ^{a} \:{F} ^{b} \:{M{L} ^{-1}} ^{c}}$

$\mathsf{[{M} ^{0}L{T} ^{-1}]\:=\:{[L]}^{a} \:{[ML{T} ^{-2}]} ^{b} \:{[M{L} ^{-1}]} ^{c}}$

$\mathsf{[{M}^{0}L{T} ^{-1}]\:=\: {[L]}^{a} \:[{M} ^{b} {L} ^{b} {T} ^{-2b}]\:{[M]} ^{c}{[{L} ^{-1}]} ^{c}}$

$\mathsf{[{M}^{0}L{T} ^{-1}]\:=\: [{M} ^{b\:+\:c}] \:[{L} ^{a\:+\:b\:-\:c}] \:[{T} ^{-2b}]}$

Now, Comparing powers of M, L and T in both side,

$\mathsf{[{M}^{0}] \:=\:[{M} ^{b\:+\:c}]}$

↪️ b + c = 0

↪️ b = - c --> ( 1 )

$\mathsf{[{L}^{1}] \:=\:[{L}^{a\:+\:b\:-\:c}]}$

↪️ 1 = a + b - c

↪️ 1 = a + 2b --> ( 2 )

$\mathsf{[{T}^{-1}] \:=\:[{T}^{-2b}]}$

↪️ - 1 = - 2b

➡️ b = $\mathsf{\dfrac{1}{2}}$

Putting value of ' b ' in equation ( 2 ),

↪️ 1 = 2b + a

↪️ 1 = a + 2 $\mathsf{\dfrac{1}{2}}$

↪️ a = 1 - 1

➡️ a = 0

Putting value of ' a ' in equation ( 1 ),

↪️ b = - c

↪️ c = - b

➡️ c = - $\mathsf{\dfrac{1}{2}}$

Equating these valued of 'a', ' b', ' c ' , in equation ( i ),

$\mathsf{v\:=\:k\:{l} ^{a} \:{F} ^{b} \:{M} ^{c}}$

$\mathsf{v\:=\:k\:{l} ^{\tiny{0}} \:{F} ^{\tiny{\dfrac{1}{2}}} \:{M} ^{\tiny{\dfrac{-1}{2}}}}$

➡️ $\boxed{\mathsf{v\:=\:k\:{\sqrt{\dfrac{F}{M}}}}}$

Answer 2

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