Question

A small spherical steel ball is placed at a distance slightly away from the center of a concave mirror having radius of curvature \(250 \mathrm{~cm}\). If the ball is released, it will now move on the curved surface. What will be the periodic time of this motion? Ignore frictional force and take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$. (A) \((\pi / 4) \mathrm{s}\) (B) \(\pi \mathrm{s}\) (C) \((\pi / 2) \mathrm{s}\) (D) \(2 \pi \mathrm{s}\)

Short answer

The short answer is: The periodic time of the motion is \(\pi \mathrm{s}\) (Answer choice B).

Step-by-step solution

Determine the length of the pendulum

The length of the pendulum is equivalent to the radius of curvature of the concave mirror, which is given as \(250 \mathrm{~cm}\). Since we need to use the value of \(\mathrm{g}\) in \(\mathrm{m/s}^2\), we will convert the length \(L\) into meters. So, \(L = 2.5 \mathrm{~m}\).

Use the formula for finding the periodic time of a simple pendulum

For a simple pendulum, the periodic time \(T\) can be given by the formula: \[T = 2\pi\sqrt{\frac{L}{g}}\] where \(L\) is the length (radius of curvature) in meters and \(g\) is the acceleration due to gravity in \(\mathrm{m/s}^2\).

Calculate the periodic time

Plug in the given values and calculate the periodic time: \[T = 2\pi\sqrt{\frac{2.5}{10}}\] \[T = 2\pi\sqrt{\frac{1}{4}}\] \[T = 2\pi\left(\frac{1}{2}\right)\] \[T = \pi \mathrm{s}\] The periodic time of the motion is \(\pi \mathrm{s}\), which corresponds to the answer choice (B).