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Question: The velocity(v) of a transverse wave on a string may depend upon (a) length l of the string, (b) ten...

The velocity(v) of a transverse wave on a string may depend upon (a) length l of the string, (b) tension T in string and (c) mass per unit length μ\mu of string. Derive the formula dimensionally.

Explanation

Solution

As a very first step, one could read the question well and then note down the given quantities and then recall the dimensions of each of these quantities. Now equate the dimension of velocity to the product of the dimensions of other quantities each raised to certain power. Then, apply the principle of homogeneity to find the final expression.

Complete step by step solution:
In the question, we are given that the velocity of a transverse wave on a string will depend upon the length of the string, tension in the string and also on the mass per unit length of the string. We are supposed to find the formula dimensionally.
As a very first step, one could recall the dimensions of each of the given quantities.
Tension on the string, [T]=[MLT−2]\left[ T \right]=\left[ ML{{T}^{-2}} \right]
Mass per unit length, [μ]=[ML−1]\left[ \mu \right]=\left[ M{{L}^{-1}} \right]
Length of the string, [l]=[L]\left[ l \right]=\left[ L \right]
Velocity of a transverse wave on a string, [v]=[LT−1]\left[ v \right]=\left[ L{{T}^{-1}} \right]
Now we could dimensional analysis as per requirement,
[v]=LT−1=[T]a[μ]b[L]c\left[ v \right]=L{{T}^{-1}}={{\left[ T \right]}^{a}}{{\left[ \mu \right]}^{b}}{{\left[ L \right]}^{c}}……………………………………………. (A)
⇒LT−1=(MaLaT−2a)(MbL−b)(Lc)\Rightarrow L{{T}^{-1}}=\left( {{M}^{a}}{{L}^{a}}{{T}^{-2a}} \right)\left( {{M}^{b}}{{L}^{-b}} \right)\left( {{L}^{c}} \right)
Now from the principle of homogeneity, we could say that,
For the mass term,
a+b=0a+b=0…………………………….. (1)
For the length term,
a−b+c=1a-b+c=1……………………………… (2)
From the time term,
−2a=−1-2a=-1
⇒a=12\Rightarrow a=\dfrac{1}{2}……………………………………. (3)
Put (3) in (1) to get,
b=−12b=-\dfrac{1}{2}……………………………………. (4)
Now put (3) and (4) in (2) to get,
c=0c=0…………………………………………. (5)
Now putting (3), (4) and (5) in equation (A), we get,
[v]=LT−1=[T]12[μ]−12[L]0\left[ v \right]=L{{T}^{-1}}={{\left[ T \right]}^{\dfrac{1}{2}}}{{\left[ \mu \right]}^{-\dfrac{1}{2}}}{{\left[ L \right]}^{0}}
Now using the method of dimensional analysis we found the expression for velocity of the wave to be given by,
v=Tμv=\sqrt{\dfrac{T}{\mu }}

Note: The major drawback of dimensionally finding the expression for physical quantities is that the constant present in these relations couldn’t be found by this method. The statement of the principle of homogeneity goes like this, ‘the dimensions of each of the terms in a dimensional equation on both sides should be the same’.