Question

A stone is swinging in a horizontal circle 0.8m in diameter at 30 $\dfrac{{rev}}{{\min }}$. A distant horizontal light beam causes a shadow of the stone to be formed on a really vertical wall. The amplitude and period of the simple harmonic motion for the shadow of the stone are
$A.$ 0.4m, 4s
$B.$ 0.2m, 2s
$C.$ 0.4m, 2s
$D.$ 0.8m, 2s

Answer 1

Time Period-
The Time Period is the time it takes for an object to complete one full cycle of its periodic motion, such as the time it takes a pendulum to make one full back and forth swing. The formula for calculating Time Period in simple harmonic motion is given by- $T = 2\pi \sqrt {\dfrac{m}{k}}$ or $T = \dfrac{1}{\upsilon }$, where $\upsilon$ is the frequency.
Amplitude- The amplitude is simply the maximum displacement of the object from the equilibrium position.

Complete answer:
Now moving to the question-
We have, diameter of the circle = 0.8m
Radius of the circle is given by = $\dfrac{d}{2} = \dfrac{{0.8}}{2} = 0.4$m
Frequency of stone swinging $= 30\dfrac{{rev}}{{\min }} = \dfrac{{30}}{{60}} = \dfrac{1}{2} = 0.5\dfrac{{rev}}{\operatorname{s} }$
Now for Time Period = $T = \dfrac{1}{\upsilon } = \dfrac{1}{{0.5}} = 2s$
For Amplitude(A) = maximum displacement = radius of circle = 0.4m
Therefore, the time period of the stone is 2s and the amplitude of the stone is given by 0.4m. The correct option for the above question is $C.$

NOTE - Simple Harmonic Motion
Simple Harmonic Motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object’s displacement magnitude and acts towards the object’s equilibrium position. It results in an oscillation which, if uninhibited by friction or any other dissipation of energy, continues indefinitely. In other words, the motion of a particle moving along a straight line with an acceleration whose direction is always towards a fixed point on the line and whose magnitude is proportional to the distance from the fixed point is called simple harmonic motion (SHM).