Sure, let's work through this problem step-by-step:
1. Recalling the formula for the period of a pendulum:
The period [tex]\( T \)[/tex] of a pendulum is given by the formula:
[tex]\[
T = 2 \pi \sqrt{\frac{L}{g}}
\][/tex]
where:
- [tex]\( T \)[/tex] is the period of the pendulum.
- [tex]\( L \)[/tex] is the length of the pendulum.
- [tex]\( g \)[/tex] is the acceleration due to gravity.
2. Given Values:
- At sea level:
[tex]\[
g_{\text{sea}} = 9.800 \, \text{m/s}^2
\][/tex]
[tex]\[
T_{\text{sea}} = 1.000 \, \text{s}
\][/tex]
- At the top of the mountain:
[tex]\[
T_{\text{mountain}} = 0.9990 \, \text{s}
\][/tex]
3. Establishing the Ratio:
Since the length [tex]\( L \)[/tex] of the pendulum is constant, we can establish the following ratio:
[tex]\[
\left( \frac{T_{\text{mountain}}}{T_{\text{sea}}} \right)^2 = \frac{g_{\text{sea}}}{g_{\text{mountain}}}
\][/tex]
4. Calculating the Ratio:
Substitute the given periods into the ratio:
[tex]\[
\left( \frac{0.9990 \, \text{s}}{1.000 \, \text{s}} \right)^2 = \left(0.9990\right)^2 = 0.998001
\][/tex]
5. Solving for [tex]\( g_{\text{mountain}} \)[/tex]:
Rearrange the equation to solve for [tex]\( g_{\text{mountain}} \)[/tex]:
[tex]\[
g_{\text{mountain}} = \frac{g_{\text{sea}}}{\text{ratio}}
\][/tex]
Substitute the values:
[tex]\[
g_{\text{mountain}} = \frac{9.800 \, \text{m/s}^2}{0.998001}
\][/tex]
[tex]\[
g_{\text{mountain}} = 9.81962943924906 \, \text{m/s}^2
\][/tex]
6. Rounding to Significant Figures:
We are asked to keep 4 significant figures, so the result is:
[tex]\[
g_{\text{mountain}} = 9.820 \, \text{m/s}^2
\][/tex]
So, the acceleration due to gravity at the top of the mountain is [tex]\( 9.820 \, \text{m/s}^2 \)[/tex].
