Question

If the time period of a two meter long simple pendulum is $2s$, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

• A. $2{\pi }^{2}m{s}^{-2}$
• B. $16m{s}^{-2}$
• C. $9.8m{s}^{-2}$
• D. ${\pi }^{2}m{s}^{-2}$

Answer 1

The correct option is A

$2{\pi }^{2}m{s}^{-2}$

Step 1: Given

Length of the pendulum: $l=2m$

Time period of the pendulum: $T=2s$

Step 2: Formula Used

$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$

Where g is the acceleration due to gravity,

Step 3: Calculate the acceleration due to gravity where SHM is executed using the formula

$$\begin{gathered} 2=2\mathrm{\pi }\sqrt{\frac{2}{\mathrm{g}}} \\ \Rightarrow 4={\left(2\mathrm{\pi }\right)}^2\frac{2}{\mathrm{g}} \\ \Rightarrow \mathrm{g}=\frac{8{\mathrm{\pi }}^2}{4} \\ \Rightarrow \mathrm{g}=2{\mathrm{\pi }}^2 \end{gathered}$$

Hence, the acceleration due to gravity at the place where the pendulum is executing S.H.M. is $2{\mathrm{\pi }}^{2}m{s}^{-2}$.