At sea level, at a latitude where , a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m. What is the value of g in m/s² at a location where the length of such a pendulum is 0.970 m? A)0.098 3 B)3.05 C)9.57 D)10.0 E)38.3

Question

Quizplus · Accepted Answer

The period of a pendulum is given by $T=2\pi\sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

At the first location, we have $T_1=2.00	ext{ s}$ and $L_1 = 0.993	ext{ m}$, so we can solve for $g$:

\begin{align*}
T_1 &= 2\pi\sqrt{\frac{L_1}{g}} \
2.00	ext{ s} &= 2\pi\sqrt{\frac{0.993	ext{ m}}{g}} \
g &= \frac{4\pi^2(0.993	ext{ m})}{4	ext{ s}^2} \
g &= 9.82	ext{ m/s}^2
\end{align*}

At the second location, we have $L_2 = 0.970	ext{ m}$ and we want to find $g_2$ such that $T_2=2.00	ext{ s}$. Using the formula for the period of a pendulum, we can solve for $g_2$:

\begin{align*}
T_2 &= 2\pi\sqrt{\frac{L_2}{g_2}} \
2.00	ext{ s} &= 2\pi\sqrt{\frac{0.970	ext{ m}}{g_2}} \
g_2 &= \frac{4\pi^2(0.970	ext{ m})}{4	ext{ s}^2} \
g_2 &= 9.57	ext{ m/s}^2
\end{align*}

Therefore, the value of $g$ at the second location is 9.57 m/s$^2$. The closest answer choice is (C) 9.57.

At Sea Level, at a Latitude Where , a Pendulum