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Answer :
To find the length of the pendulum rope on Mars, we use the relationship between the frequency of a pendulum and its length. The formula for the frequency [tex]\( f \)[/tex] of a simple pendulum is given by:
[tex]\[
f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}
\][/tex]
Where:
- [tex]\( f \)[/tex] is the frequency of the pendulum,
- [tex]\( g \)[/tex] is the acceleration due to gravity,
- [tex]\( L \)[/tex] is the length of the pendulum,
- [tex]\(\pi\)[/tex] is a constant approximately equal to 3.14159.
We are trying to find the length [tex]\( L \)[/tex], so we need to rearrange the formula to solve for [tex]\( L \)[/tex]:
1. Start with the formula for frequency:
[tex]\[
f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}
\][/tex]
2. Square both sides to eliminate the square root:
[tex]\[
f^2 = \left(\frac{1}{2\pi}\right)^2 \left(\frac{g}{L}\right)
\][/tex]
3. Simplify:
[tex]\[
f^2 = \frac{g}{4\pi^2 L}
\][/tex]
4. Solve for [tex]\( L \)[/tex]:
[tex]\[
L = \frac{g}{4\pi^2 f^2}
\][/tex]
Given:
- The frequency [tex]\( f \)[/tex] is 35.9 Hz.
- The gravitational acceleration [tex]\( g \)[/tex] on Mars is 3.73 m/s².
Substitute these values into the formula:
[tex]\[
L = \frac{3.73}{4\pi^2 \cdot (35.9)^2}
\][/tex]
When you calculate this, the length [tex]\( L \)[/tex] of the pendulum rope is approximately:
[tex]\[
7.33 \times 10^{-5} \text{ meters}
\][/tex]
This value represents the length of the rope needed for the pendulum to have a frequency of 35.9 Hz under the gravitational conditions on Mars.
[tex]\[
f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}
\][/tex]
Where:
- [tex]\( f \)[/tex] is the frequency of the pendulum,
- [tex]\( g \)[/tex] is the acceleration due to gravity,
- [tex]\( L \)[/tex] is the length of the pendulum,
- [tex]\(\pi\)[/tex] is a constant approximately equal to 3.14159.
We are trying to find the length [tex]\( L \)[/tex], so we need to rearrange the formula to solve for [tex]\( L \)[/tex]:
1. Start with the formula for frequency:
[tex]\[
f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}
\][/tex]
2. Square both sides to eliminate the square root:
[tex]\[
f^2 = \left(\frac{1}{2\pi}\right)^2 \left(\frac{g}{L}\right)
\][/tex]
3. Simplify:
[tex]\[
f^2 = \frac{g}{4\pi^2 L}
\][/tex]
4. Solve for [tex]\( L \)[/tex]:
[tex]\[
L = \frac{g}{4\pi^2 f^2}
\][/tex]
Given:
- The frequency [tex]\( f \)[/tex] is 35.9 Hz.
- The gravitational acceleration [tex]\( g \)[/tex] on Mars is 3.73 m/s².
Substitute these values into the formula:
[tex]\[
L = \frac{3.73}{4\pi^2 \cdot (35.9)^2}
\][/tex]
When you calculate this, the length [tex]\( L \)[/tex] of the pendulum rope is approximately:
[tex]\[
7.33 \times 10^{-5} \text{ meters}
\][/tex]
This value represents the length of the rope needed for the pendulum to have a frequency of 35.9 Hz under the gravitational conditions on Mars.
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