Solution
This problem can be solved using dimensional analysis and the scaling properties of motion under gravity.
For a particle sliding under gravity on a fixed path, the time period of oscillation $T$ depends on the length scale of the path ($L$) and the acceleration due to gravity ($g$). The dimensional formula for time is $[T]$.
- Dimension of Length $[L] = L$
- Dimension of Acceleration $[g] = L T^{-2}$
The only combination of $L$ and $g$ that yields the dimension of time is:
$$ T \propto \sqrt{\frac{L}{g}} $$The frequency of oscillation $f$ is the reciprocal of the time period ($f = 1/T$). Therefore:
$$ f \propto \sqrt{\frac{g}{L}} $$This implies that frequency is inversely proportional to the square root of the linear dimensions.
We are given that the linear dimensions of the frame are made four times larger. Let the initial length be $L$ and the new length be $L’$.
$$ L’ = 4L $$Let the initial frequency be $f$ and the new frequency be $f’$. Using the proportionality found above:
$$ \frac{f’}{f} = \sqrt{\frac{L}{L’}} = \sqrt{\frac{L}{4L}} = \sqrt{\frac{1}{4}} = \frac{1}{2} $$ $$ f’ = \frac{1}{2} f $$The frequency of the periodic motion changes by a factor of $1/2$.