Question

A pendulum having a time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a seconds pendulum at a place where $g = {\pi ^2}m/{s^2}$.

Answer 1

The above problem can be resolved by using the fundamental relation for the time period of the pendulum. In the formula of the time period for the simple pendulum, the formula's variables are generally the length of the pendulum and the value of gravitational acceleration. The problem is given with the value of the time, and upon substituting this, we will get a mathematical relation. Then on squaring this relation, somebody can obtain the desired value of the length of the pendulum.

Complete step by step answer:
Given:
The time period of the seconds pendulum is, $T = 2\;{\rm{s}}$.
The gravitational acceleration is given as, $g = {\pi ^2}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$.
Let us assume that L is the length of the seconds pendulum.
Then, the expression for the time period of second pendulum is given as,
$T = 2\pi \sqrt {\dfrac{L}{g}}$
Solve by substituting the values in above expression as,
$2\;{\rm{s}} = 2\pi \sqrt {\dfrac{L}{{{\pi ^2}}}}$
Squaring both sides of above equation as,

{\left( {2\;{\rm{s}}} \right)^2} = {\left( {2\pi \sqrt {\dfrac{L}{{{\pi ^2}}}} } \right)^2}\\\ 4 = 4{\pi ^2} \times \dfrac{L}{{{\pi ^2}}}\\\ L = 1\;{\rm{m}} \end{array}$$ Therefore, the length of a seconds pendulum at a place is $$1\;{\rm{m}}$$. **Note:** To resolve the above problem, it is necessary to remember the formula for the time period and the various significant variables like length of the pendulum and gravitational acceleration. The time period of the simple pendulum varies inversely with the gravitational acceleration and varies directly with the pendulum's length, taken into consideration. Moreover, it is also necessary to remember the fundamental concept of the period's variation with other physical quantities present around.