a simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. the…

a simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. the pendulum has a length of 2 m and its period is measured to be 2 s. the value of g obtained in this investigation is most nearly\na 1 m/s2\nb 2 m/s2\nc 5 m/s2\nd 10 m/s2\ne 20 m/s2

a simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. the pendulum has a length of 2 m and its period is measured to be 2 s. the value of g obtained in this investigation is most nearly\na 1 m/s2\nb 2 m/s2\nc 5 m/s2\nd 10 m/s2\ne 20 m/s2

Answer

Explanation:

Step1: Recall the formula for the period of a simple - pendulum

The formula for the period $T$ of a simple pendulum is $T = 2\pi\sqrt{\frac{l}{g}}$, where $T$ is the period, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity. We need to solve this formula for $g$. First, square both sides of the equation: $T^{2}=4\pi^{2}\frac{l}{g}$. Then, cross - multiply to get $gT^{2}=4\pi^{2}l$. Finally, solve for $g$: $g=\frac{4\pi^{2}l}{T^{2}}$.

Step2: Substitute the given values

We are given that $l = 2m$ and $T = 2s$. Substitute these values into the formula for $g$: $g=\frac{4\pi^{2}\times2}{2^{2}}$. Since $\pi^{2}\approx9.87$, then $4\pi^{2}\times2=4\times9.87\times2 = 78.96$ and $2^{2}=4$. So $g=\frac{78.96}{4}=19.74\approx20m/s^{2}$.

Answer:

E. $20m/s^{2}$