Thank you for visiting The frequency of a pendulum is measured to be 35 9 Hz If this pendulum is found on Mars with an acceleration due to gravity. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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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