A pendulum clock can be approximated as a simple pendulum of length 1.14 m and keeps accurate time at a location where g = 9.90 m/s2. In a location where g = 9.75 m/s2, what must be the new length of the pendulum, such that the clock continues to keep accurate time (that is, its period remains the same)?

2. Originally Posted by lovinhockey26
A pendulum clock can be approximated as a simple pendulum of length 1.14 m and keeps accurate time at a location where g = 9.90 m/s2. In a location where g = 9.75 m/s2, what must be the new length of the pendulum, such that the clock continues to keep accurate time (that is, its period remains the same)?
if i remember it right, $\displaystyle T= 2\pi \sqrt{\frac{L}{g}}$ where $\displaystyle L$ is the length of the pendulum, $\displaystyle g$ is the gravity and $\displaystyle T$ is the period.

thus, if you want to solve for the length of the pendulum that will yield the same period at different gravity,

$\displaystyle 2\pi \sqrt{\frac{L_1}{g_1}} = 2\pi \sqrt{\frac{L_2}{g_2}}$
or $\displaystyle \sqrt{\frac{L_1}{g_1}} = \sqrt{\frac{L_2}{g_2}}$
or just $\displaystyle \frac{L_1}{g_1} = \frac{L_2}{g_2}$.. and just plug in the values correspondingly..