Question

A simple pendulum of length $1m$ is oscillating with an angular frequency $10rad/s$ . The support of the pendulum starts oscillating up and down with a small angular frequency of $1rad/s$ and an amplitude of ${{10}^{-2}}m$ .The relative change in the angular frequency of the pendulum is best given by ?

Answer 1

Hint: Since the support of the pendulum is also oscillating up and down with a small angular frequency, the net gravitational force on the pendulum block will change from $g$ to ${{g}_{eff}}$ . Thus, we will apply the concept of relative change for a variable to calculate the relative change in frequency of the pendulum.

Complete answer:
We know,
$\Rightarrow \omega =\sqrt{\dfrac{{{g}_{eff}}}{l}}$
Taking log to the base e on both sides, our equation becomes:
$\Rightarrow \ln (\omega )=\dfrac{1}{2}\left[ \ln ({{g}_{eff}})-\ln (l) \right]$
On differentiating both sides, we get:
$\Rightarrow \dfrac{d\omega }{\omega }=\dfrac{1}{2}\dfrac{d{{g}_{eff}}}{{{g}_{eff}}}-\dfrac{1}{2}\dfrac{dl}{l}$
Since, there is no change in the length of pendulum, the term: $\dfrac{dl}{l}=0$
Therefore, our final equation is:
$\Rightarrow \dfrac{d\omega }{\omega }=\dfrac{1}{2}\dfrac{d{{g}_{eff}}}{{{g}_{eff}}}$
For a finite change, this can be written as:
$\Rightarrow \dfrac{\vartriangle \omega }{\omega }=\dfrac{1}{2}\dfrac{\vartriangle {{g}_{eff}}}{{{g}_{eff}}}$ [Let this be equation number 1]
Now in the above equation, it is given:
$\Rightarrow \omega =10rad/s$
$\Rightarrow {{g}_{eff}}=g$
And $\vartriangle {{g}_{eff}}$ can be calculated as follows:
Since, the support is also oscillating it will exert a centrifugal force on the pendulum. Therefore,
The net change in ${{g}_{eff}}$ will be:
$\begin{align}
  & \Rightarrow \vartriangle {{g}_{eff}}={{\omega }_{s}}^{2}A-(-{{\omega }_{s}}^{2}A) \\
 & \Rightarrow \vartriangle {{g}_{eff}}=2{{\omega }_{s}}^{2}A \\
\end{align}$
Where,
${{\omega }_{s}}$ is the angular frequency of the support
$A$ is the amplitude of support
And the value of these terms is given as:
$\begin{align}
  & \Rightarrow {{\omega }_{s}}=1rad/s \\
 & \Rightarrow A={{10}^{-2}}m \\
\end{align}$
Therefore,
$\begin{align}
  & \Rightarrow \vartriangle {{g}_{eff}}=2\times {{(1)}^{2}}\times {{10}^{-2}} \\
 & \Rightarrow \vartriangle {{g}_{eff}}=2\times {{10}^{-2}}m/{{s}^{2}} \\
\end{align}$
Now, putting the values of $\omega ,{{g}_{eff}}$ and $\vartriangle {{g}_{eff}}$ in equation number (1), we get:
Relative change in angular frequency:
$\begin{align}
  & \Rightarrow \dfrac{\vartriangle \omega }{\omega }=\dfrac{1}{2}\left( \dfrac{2\times {{10}^{-2}}}{10} \right) \\
 & \Rightarrow \dfrac{\vartriangle \omega }{\omega }={{10}^{-3}} \\
\end{align}$
Hence, the relative change in angular frequency comes out to be ${{10}^{-3}}$ .

Note:
While calculating relative change, all the terms inside the logarithmic are opened as it is, i.e., positive terms are positive and negative terms are negative. But, when calculating for “Errors”, all the terms should be taken positively as we should always find the maximum possible error. Also, relative change is a dimensionless quantity.