Question

for the following exercise, use the model for the period of a pendulum, $t$, such that $t = 2pisqrt{\frac{l}{g}}$, where the length of the pendulum is $l$ and the acceleration due to gravity is $g$. if the acceleration due to gravity is $9.8 m/s^{2}$ and the period equals $5 s$, find the length to the nearest cm ($100 cm = 1 m$).

Explanation:

Step1: Rearrange the pendulum - period formula for $L$.

Given $T = 2\pi\sqrt{\frac{L}{g}}$, first square both sides: $T^{2}=(2\pi)^{2}\frac{L}{g}$. Then solve for $L$: $L=\frac{T^{2}g}{4\pi^{2}}$.

Step2: Substitute the given values of $T$ and $g$ into the formula for $L$.

We know that $T = 5s$ and $g=9.8m/s^{2}$. Substitute these values into $L=\frac{T^{2}g}{4\pi^{2}}$:
\[

$$\begin{align*} L&=\frac{5^{2}\times9.8}{4\pi^{2}}\\ &=\frac{25\times9.8}{4\pi^{2}}\\ &=\frac{245}{4\pi^{2}} \end{align*}$$

\]
\[

$$\begin{align*} L&=\frac{245}{4\times3.14^{2}}\\ &=\frac{245}{4\times9.8596}\\ &=\frac{245}{39.4384}\\ &\approx6.21m \end{align*}$$

\]

Step3: Convert the length from meters to centimeters.

Since $1m = 100cm$, then $L = 6.21m=6.21\times100 = 621cm$.

Answer:

$621$