Question

The frequency $\mathrm{n}$ of vibration of a stretched string depends upon its length $\mathrm{l}$, its mass per unit length $\mathrm{m}$ and Tension $\mathrm{t}$ in the string. Obtain dimensionally an expression for frequency $\mathrm{n}$.

Answer 1

Step-1: Given data

As per question the frequency $\mathrm{n}$ of vibration depends on $\mathrm{l},\mathrm{m},\mathrm{t}$ .

As $\mathrm{l}$is length so its dimension will be: $\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]$

$\mathrm{m}$ is mass per unit length so: $\left[{\mathrm{M}}^1{\mathrm{L}}^{-1}{\mathrm{T}}^0\right]$

$\mathrm{t}$ is a tension which is a form of force. so: $\left[{\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]$

Step-2: Calculate the dimension of $\mathrm{n}$

As the dimension depends on $\mathrm{m},\mathrm{l}\mathrm{and}\mathrm{t}$ .

So, let the dimension of $\mathrm{n}$ be $\mathrm{n}=\mathrm{K}\left[{\mathrm{m}}^{\mathrm{a}}{\mathrm{l}}^{\mathrm{b}}{\mathrm{t}}^{\mathrm{c}}\right]$; here K is a dimensional constant.

Actual dimension of $\mathrm{n}$ = $\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{-1}\right]$

Substituting the values in $\mathrm{n}$ and equating.

$$\begin{gathered} \left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{-1}\right]=\mathrm{K}{\left[{\mathrm{M}}^1{\mathrm{L}}^{-1}{\mathrm{T}}^0\right]}^{\mathrm{a}}{\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]}^{\mathrm{b}}{\left[{\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]}^{\mathrm{c}} \\ \Rightarrow \left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{-1}\right]=\mathrm{K}\left[{\mathrm{M}}^{\mathrm{a}}{\mathrm{L}}^{-\mathrm{a}}{\mathrm{T}}^0\right]\left[{\mathrm{M}}^0{\mathrm{L}}^{\mathrm{b}}{\mathrm{T}}^0\right]\left[{\mathrm{M}}^{\mathrm{c}}{\mathrm{L}}^{\mathrm{c}}{\mathrm{T}}^{-2\mathrm{c}}\right] \\ \Rightarrow \left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{-1}\right]=\mathrm{K}[{\mathrm{M}}^{\mathrm{a}+\mathrm{c}}{\mathrm{L}}^{-\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{T}}^{-2\mathrm{c}}] \end{gathered}$$

Comparing LHS and RHS

$$\begin{gathered} -2\mathrm{c}=-1 \\ \Rightarrow \mathrm{c}=\frac{1}{2} \end{gathered}$$ and $$\begin{gathered} \mathrm{a}+\mathrm{c}=0 \\ \Rightarrow \mathrm{a}=-\frac{1}{2} \end{gathered}$$ and $$\begin{gathered} -\mathrm{a}+\mathrm{b}+\mathrm{c}=0 \\ \Rightarrow \mathrm{b}=\mathrm{a}-\mathrm{c} \\ \Rightarrow \mathrm{b}=-\frac{1}{2}-\frac{1}{2}=-1 \end{gathered}$$

So the dimension of $\mathrm{n}$ becomes $\left[\mathrm{n}\right]=\mathrm{K}[{\mathrm{m}}^{\frac{-1}{2}}{\mathrm{l}}^{-1}{\mathrm{t}}^{\frac{1}{2}}]$

The possible expression of $\mathrm{n}$is $n=K.\frac{1}{l}\sqrt{\frac{t}{m}}$