Question

velocity of sound depends on density and modulus of elasticity [E].the dimensional formula of E is [ML^-1T^-2]. Using the principle of homogeneity of dimensions ,arrive at an expression for the velocity of sound (Take K=1)​

Answer 1

Answer :-

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Dimensional Formula had been used. We are given that the Velocity of Sound waves depend on the density of medium and modulus of Elasticity. This means that Velocity of Sound Waves depend on Density and Elasticity which are raised to some power. First we will find the main equation of Velocity. Then we will different dimensional quantities and find the answer.

Let's do it !!

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★ Solution :-

Given,

» Velocity of sound waves depend on modulus of Elasticity and Density of medium.

» K = 1

» Dimension of Mass = [M]

» Dimension of Time = [T]

» Dimension of Length = [L]

• Let the velocity of sound waves be 'v'

• Let the Modulus of Elasticity be 'E'

• Let the Density of Medium be 'd'

Then,

$\\\;\;\;\sf{:\mapsto\;\;\;v\;\;\propto\;\;\bf{d^{a}\;\;E^{b}}}$

Here a and b are some exponents by which d and E should be multiplied to get their original values respectively. Then,

$\\\;\;\;\sf{:\mapsto\;\;\;v\;\;=\;\;\bf{K\;\;d^{a}\;\;E^{b}}}$

where K = a dimensionless constant.

Let this be equation (i).

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~ For Dimensions of 'v', 'd' and 'E' ::

We know that,

$\\\;\;\sf{:\rightarrow\;\;Density,\;d\;=\;\bf{\dfrac{Mass}{Volume}}}$

$\\\;\;\sf{:\rightarrow\;\;Density,\;d\;=\;\bf{\dfrac{[M^{1}]}{[L^{3}]}}}$

$\\\;\;\bf{:\Rightarrow\;\;Density,\;d\;=\;\bf{[M^{1}\:\:L^{-3}]}}$

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$\\\;\;\sf{:\rightarrow\;\;Velocity,\;v\;=\;\bf{\dfrac{Length}{Time}}}$

$\\\;\;\sf{:\rightarrow\;\;Velocity,\;v\;=\;\bf{\dfrac{[L^{1}]}{[T^{1}]}}}$

$\\\;\;\bf{:\Rightarrow\;\;Velocity,\;v\;=\;\bf{[L^{1}\:T^{-1}]}}$

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Also, its given that,

$\\\;\;\bf{:\Rightarrow\;\;Modulus\;of\;Elasticity,\;E\;=\;\bf{[M^{1}\:L^{-1}\:T^{-2}]}}$

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~ For Formula of Velocity of Sound ::

Now by Dimensional Method of Analysis we know that, a equation is correct if the dimension of LHS is equal to dimension of RHS.

Let's apply these values in the equation (i), and find our answer. Then,

$\\\;\;\;\sf{:\Longrightarrow\;\;\;v\;\;=\;\;\bf{K\;\;d^{a}\;\;E^{b}}}$

$\\\;\;\;\sf{:\Longrightarrow\;\;\;[L^{1}\:T^{-1}]\;\;=\;\;\bf{K\;\;[M\:L^{-3}]^{a}\;\;[M\:L^{-1}\:T^{-2}]^{b}}}$

Since, K = 1

$\\\;\;\;\sf{:\Longrightarrow\;\;\;[L^{1}\:T^{-1}]\;\;=\;\;\bf{[M\:L^{-3}]^{a}\;\;[M\:L^{-1}\:T^{-2}]^{b}}}$

Now by making the value of LHS and RHS similar and common, we get,

$\\\;\;\;\sf{:\Longrightarrow\;\;\;[M^{0}\:L^{1}\:T^{-1}]\;\;=\;\;\bf{M^{a\;+\;b}\;L^{-3a\;-\;b}\;T^{-2b}}}$

Now comparing the value of M, L and T from LHS and RHS, we get,

✒ a + b = 0 , -3a - b = 1 , -2b = -1

On solving this,

✒ a = -b , 3a = 1 + b , 2b = 1

$\\\;\;\sf{\rightarrow\;\;\;\;a\;=\;\dfrac{-1}{2}\;,\;\;\;\;b\;=\;\dfrac{1}{2}}$

By applying these values, we get,

$\\\;\;\;\sf{:\mapsto\;\;\;v\;\;=\;\;\bf{K\;\;d^{(-1)/2}\;\;E^{1/2}}}$

$\\\;\;\;\sf{:\mapsto\;\;\;v\;\;=\;\;\bf{1\;\;\bigg(\dfrac{1}{d}\bigg)^{1/2}\;\;E^{1/2}}}$

$\\\;\;\;\bf{:\mapsto\;\;\;v\;\;=\;\;\bf{\sqrt{\dfrac{E}{d}}}}$

Hence, we got the expression.

$\\\;\;\underline{\boxed{\rm{Velocity\;of\;Sound,\;v\;=\;\bf{\sqrt{\dfrac{E}{d}}}}}}$

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★ More to know :-

• Dimensions of Electric Current = [A]

• Dimensions of Temperature = [K]

• Dimensions of Luminous Intensity = [cd]

• Dimensions of amount of substance = [mol]

• Derivation of Dimensions of Modulus of Elasticity :-

$\\\;\;\tt{\leadsto\;\;\;E\;\;=\;\;\dfrac{Stress}{Strain}}$

$\\\;\;\tt{\leadsto\;\;\;E\;\;=\;\;\dfrac{Force}{Area\;\times\;Strain}}$

Strain is dimensionless quantity.

$\\\;\;\tt{\leadsto\;\;\;E\;\;=\;\;\dfrac{Mass\;\times\;Acceleration}{Area\;\times\;1}}$

$\\\;\;\tt{\leadsto\;\;\;E\;\;=\;\;\dfrac{M\;L\;T^{-2}}{L^{2}}}$

$\\\;\;\tt{\leadsto\;\;\;E\;\;=\;\;\dfrac{M\;T^{-2}}{L^{1}}}$

$\\\;\;\tt{\leadsto\;\;\;E\;\;=\;\;M\;L^{-1}\;T^{-2}}$

$\\\;\;\tt{\leadsto\;\;\;[E]\;\;=\;\;[M\;L^{-1}\;T^{-2}]}$

Hence, derived.