Question

At a place where \[g = 980\,{\text{cm/}}{{\text{s}}^2}\]. The length of second’s pendulum is about:
A. \[2\,{\text{cm}}\]
B. \[2\,{\text{m}}\]
C. \[100\,{\text{cm}}\]
D. \[200\,{\text{cm}}\]

Answer 1

Hint:Use the formula for time period of a simple pendulum. This formula gives the relation between time period of simple pendulum, length of the simple pendulum and acceleration due to gravity. Recall the concept of the second's pendulum and time period of this second’s pendulum. Substitute this value in the formula and determine the length of the second’s pendulum.

Formula used:
The time period \[T\] of a simple pendulum is given by
\[T = 2\pi \sqrt {\dfrac{L}{g}} \] …… (1)
Here, \[L\] is the length of the simple pendulum and \[g\] is acceleration due to gravity.

Complete step by step answer:
We have given that the value of acceleration due to gravity is \[980\,{\text{cm/}}{{\text{s}}^2}\].
\[g = 980\,{\text{cm/}}{{\text{s}}^2}\]
We have asked to determine the length of a second’s pendulum. A second’s pendulum is a simple pendulum having a time period of two seconds. One second out of these two seconds is required for one swing of the pendulum. Hence, the time period of the second’s pendulum is two seconds.
\[T = 2\,{\text{s}}\]
Rearrange equation (1) for length of the second’s pendulum.
\[L = \dfrac{{g{T^2}}}{{4\pi }}\]
Substitute \[980\,{\text{cm/}}{{\text{s}}^2}\] for \[g\], \[2\,{\text{s}}\] for \[T\] and \[3.14\] for \[\pi \] in the above equation.
\[L = \dfrac{{\left( {980\,{\text{cm/}}{{\text{s}}^2}} \right){{\left( {2\,{\text{s}}} \right)}^2}}}{{4{{\left( {3.14} \right)}^2}}}\]
\[ \Rightarrow L = 99.39\,{\text{cm}}\]
\[ \therefore L \approx 100\,{\text{cm}}\]
Therefore, the length of the second’s pendulum is \[100\,{\text{cm}}\].

Hence, the correct option is C.

Note: The students should keep in mind that there is no need to convert the unit of acceleration due to gravity from the CGS system of units to the SI system of units as the final answer for the length of the simple pendulum is in the CGS system of units. Hence, the extra efforts to convert the unit of acceleration due to gravity will be reduced.