a 1.25 m long pendulum on mars has a period of 3.65 s. what is the acceleration of gravity g on mars? (unit…

Explanation:

Step1: Recall the pendulum - period formula

The period formula for 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 $g$.

Step2: Square both sides of the formula

Starting from $T = 2\pi\sqrt{\frac{L}{g}}$, we square both sides: $T^{2}=(2\pi)^{2}\frac{L}{g}$. Then $T^{2}=4\pi^{2}\frac{L}{g}$.

Step3: Rearrange the formula to solve for $g$

Cross - multiply to get $gT^{2}=4\pi^{2}L$, and then $g=\frac{4\pi^{2}L}{T^{2}}$.

Step4: Substitute the given values

We are given that $L = 1.25$ m and $T = 3.65$ s. Substitute these values into the formula: [ \begin{align*} g&=\frac{4\pi^{2}\times1.25}{3.65^{2}}\ &=\frac{4\times9.87\times1.25}{13.3225}\ &=\frac{49.35}{13.3225}\ &\approx3.70\text{ m/s}^2 \end{align*} ]

Answer:

$3.70$ m/s²