Question

If the radius of earth is shrinking by 0.2% without change in its mass. The time period of oscillation of a simple pendulum will be
A. Increases by \[0.2\%\]
B. Decreases by \[0.2\%\]
C. Increase by \[0.1\%\]
D. Decreases by \[0.1\%\]

Answer 1

Hint: in order to solve this problem, first of all the time period should be known. It is the time taken to complete a full cycle of oscillation. This time period equation is used up here to get the relationship between radius of earth and time period.

Complete step by step answer:
The time period for an oscillation of a simple pendulum is given by the formula
\[T=2\pi \sqrt{\dfrac{l}{g}}\]
In which \[T\] is the time period in second, \[l\] is the length of the pendulum given in metre and \[g\] is the acceleration due to gravity.
From this expression we can get to know that the time period is inversely proportional to the square root of the acceleration due to gravity.
As we all know, acceleration due to gravity is given by the formula,
\[g=\dfrac{GM}{{{R}^{2}}}\]
Where \[R\] is the radius of the Earth. Substituting this in the equation will give,
\[T=2\pi \sqrt{\dfrac{l\times {{R}^{2}}}{GM}}\]
From this expression we can see that, time period of oscillation of a simple pendulum on the surface of earth is directly proportional to the radius of earth.
Therefore
\[\dfrac{\Delta T}{T}\times 100=-\dfrac{1}{2}\times 2\times \dfrac{\Delta R}{R}\times 100=-0.2%\]
Hence it decreases with the radius of curvature and the rate will be the same as that of the radius of earth.

Therefore the correct answer is option B.

Note:
The time period which is abbreviated as T is the time taken to complete one full cycle of vibration or oscillation to pass a certain point. As the frequency of a wave gets increased, then the time period of the wave gets decreased. The unit of time period is given in 'seconds'.