Question

Show dimensionally that the frequency of transverse waves in a string of length and mass per unit length m under a tension T is given by, $n=\dfrac{k\sqrt{T}}{l\sqrt{m}}$.

Answer 1

Hint: As the question required the dimension of the frequency n so we will substitute the dimensions of T, l and m and solve the question. After solving we will come across the values of a, b and c by comparison and after substituting these in the equation $n=k{{T}^{a}}{{l}^{b}}{{m}^{c}}$ we will have our answer.

Formula used:
$\left[ T \right]=ML{{T}^{-2}},\left[ l \right]=L,\left[ m \right]=M{{L}^{-1}},\left[ n \right]={{T}^{-1}}$ where T is tension, l is length of string, m is mass and n is frequency.

Complete answer:
Suppose that n is the frequency and T is called tension l be the length of the string and m denotes the mass. We will take the relation between T, l m and v as a frequency $n=k{{T}^{a}}{{l}^{b}}{{m}^{c}}$ …(i), where k is a dimensionless constant.

As we know the dimensions of tension $\left[ T \right]=ML{{T}^{-2}}$ , dimension of $\left[ l \right]=L$ and dimension of mass is $\left[ m \right]=M{{L}^{-1}}$ and $\left[ n \right]={{T}^{-1}}$ .
Now, we will substitute all the dimensions of the terms in the equation (i). By doing so, we will get the following type equation containing dimensions.

$\begin{align}
  & {{T}^{-1}}={{\left( ML{{T}^{-2}} \right)}^{a}}{{L}^{b}}{{\left( M{{L}^{-1}} \right)}^{c}} \\
 & \Rightarrow {{T}^{-1}}={{M}^{\left( a+c \right)}}{{L}^{\left( a+b-c \right)}}{{T}^{-2a}} \\
\end{align}$

Now, we will equate the degrees of both sides of the above equation with each other by comparing the same terms. Thus, we will get $-2a=-1,a+c=0,a+b-c=0$.

After solving these equations, we get $a=\dfrac{1}{2},c=\dfrac{-1}{2},b=-1$. Now, it is time to put all the values of a, b and c in the equation $n=k{{T}^{a}}{{l}^{b}}{{m}^{c}}$. Therefore, we have

$\begin{align}
  & n=k{{T}^{a}}{{l}^{b}}{{m}^{c}} \\
 & \Rightarrow n=k{{T}^{\dfrac{1}{2}}}{{l}^{-1}}{{m}^{\dfrac{-1}{2}}} \\
 & \Rightarrow n=\dfrac{k\sqrt{T}}{l\sqrt{m}} \\
 & \Rightarrow n=k\dfrac{\sqrt{T}}{l\sqrt{m}} \\
\end{align}$

Hence, the required dimension of the frequency is $n=k\dfrac{\sqrt{T}}{l\sqrt{m}}$.

Note:
To get the dimension of the frequency we need to first equate it to its actual value, in this case $n=k{{T}^{a}}{{l}^{b}}{{m}^{c}}$ and after that the correct use of dimensions of parameters in n we will get the required answer. The most challenging part of this question is to write the correct equation of n otherwise, we will lead towards the wrong answer. One is not allowed to stop the answer at $n=k{{T}^{\dfrac{1}{2}}}{{l}^{-1}}{{m}^{\dfrac{-1}{2}}}$. We use $\sqrt{T}$ as ${{T}^{\dfrac{1}{2}}}$ only. And any term with a power of negative integer should be changed to positive by taking it to the denominator. Or if the negative power is in the denominator then, placing it to the numerator will give rise to a better solution.