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The spaceship Intergalactiea lardis OH the surface of the uninhabited Fink Planet. which orbits rather average Sar In (he' distat Garbamza Galuxy. scoUting&quo...

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The spaceship Intergalactiea lardis OH the surface of the uninhabited Fink Planet. which orbits rather average Sar In (he' distat Garbamza Galuxy. scoUting" party sts OUt do eplce. The party'$ leader physicist; ntutully-ituediately makes deterrmination Of the acceleration due to gravity ( the Fink Planet' s surlace by Means O1 simple pendulum ot length LI2 m She sets Ilee pendulum swinging; und her collaborators caretully count HO8 complete cyeles = ot oscillation during 220X

The spaceship Intergalactiea lardis OH the surface of the uninhabited Fink Planet. which orbits rather average Sar In (he' distat Garbamza Galuxy. scoUting" party sts OUt do eplce. The party'$ leader physicist; ntutully-ituediately makes deterrmination Of the acceleration due to gravity ( the Fink Planet' s surlace by Means O1 simple pendulum ot length LI2 m She sets Ilee pendulum swinging; und her collaborators caretully count HO8 complete cyeles = ot oscillation during 220X [0' What js the' result' acceleratia due [o eravity: nVs" pincyncG



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Astronauts on a distant planet set up a simple pendulum of length 1.2 $\mathrm{m}$ . The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet?

Hi there. So for this problem we are told that astronauts and set up a simple pendulum on a distant planet and the length of this pendulum is This distance L. is equal to 1.2 m were also given that the pendulum execute simple harmonic motion and mates 10 complete vibrations In 282nd. So to solve this problem, we know that the magnitude G. Of the acceleration due to grab it in, that's the one that we want to uh determine for this planet. And it's related to the length and frequency of the simple pendulum. By the following formula, we know that two pi the frequency is equal to the square root of the acceleration due to gravity over the length of the pendulum. Now squaring both sides of this, we can solve for the acceleration when we square this, we're gonna have four times p square the uh the frequency is squared, this is G over L. So we can so for G. So that is for pi square the frequency square times the length square. Now the frequency is the number of complete vibrations made per second. So in measuring the frequency of a simple pendulum on the astronauts um record And that is the complete vibrations, we're going to set any complete vibrations, we know that that is 100 And to the times that it tastes and we know that that is 280 seconds. So for the frequency we will have that this is equal to the number of complete vibrations over time. So this we can just substitute that into the equation for the acceleration and just obtained that value. So now with this we're going to attain that the acceleration due to gravity on this planet is four pi squared times the frequency square as you can see from that. So it will be the number of vibrations over the time. This squared times the length of the pendulum. So we substitute all of these values. We have four pi squared, The number of vibrations and 100 is two squared and then length, which is 1.2 m. This over the time square, 1/282, this is square. So we will find that the acceleration due to gravity on this new planet is six m/s square. So that's a solution for this problem.

All right. So for this problem, we have two pendulums ones on the planet Earth and ones on some unknown planet. And both have the same length string. However, they're on different planets, so they're experiencing different gravity's, and they both have separate periods. So the period of the pendulums on earth is 0.65 seconds in the period of the pendulums on the unknown, Planet P is equal to 0.862 seconds. And the first thing we have to answer is is the gravitational field on the unknown planet stronger or weaker than the gravitational field on Earth? So we're gonna look at the equation for the period of a pendulum is equal to pi times the square root of length of pendulum divide by the gravitational field of experiencing, and we're going to you analytically say All right, So if G is larger, then we're going to have a smaller period. Since she's in the denominator of this equation. If g of smaller, they won't have a larger period because in the nominee of the equation, there's an inverse law going on here. So then we're going to analyze the period of the pendulum on the Unknown Planet. We noticed that it's greater. It is greater than TCBY or he just read it like this mathematically. So if TCP is greater than TCBY, then Jesus Api. The gravity on the Unknown Planet must be lesser because there's an inverse law here. We state that the period of any pens alone is proportional to the inverse of the gravitational field with pendulum is experiencing. That's the first thing we ventured. We already know that G is going to be lesser on the Unknown Planet than it is gonna be on Earth simply due to this relation here. Now we just have to learn what is Guille neo known planet? Well, we can say we can compare the two periods and thusly compare the two gravity's through their equations. So let's do that. Let's say T sub e for the period of the pendulum Earth divided by the period of the pendulum on the Unknown Planet, it's simply going to be equal to their individual equations. So it's going to be two pi times the square root of the length of the pendulum. And remember, the length of the pendulum doesn't change that the same length pendulum on both Earth and the Unknown Planet divided by the gravity on earth. And this whole product is going to be divided by the equation for the period on the Unknown planet two pi times square, root of the length divided by the gravity on the Unknown Planet. All right, so what immediately cancels out here? The two pies cancel out so you can rewrite this and we can extend the square root to cover over this whole set of three fractions. If we do our fraction division correctly, we should notice that the else cancel this whole thing legal to the square root of geese API divided by gcb. All right, so because we have values for Tee Savi and TCP, we can plug them into this equation here and swear the whole equation today to get geese api Overdue city out of the square root. So it's plug in the values and then swear them. Let's say TV 0.65 This would be 0.65 divided by TCP, which is 0.862 We want to square this little thing and say that's equal to gee soapy, divided by gcb Now this is Multiply gcb over and evaluate this. If we evaluate this, it's going to be equal to 0.56 mine. And because we know the value of gcb to be 9.8, the gravity on Earth gravitational acceleration on earth 0.569 times 9.8 is equal to the gravitational acceleration on the planet peak. It's the gravitational acceleration. The Planet P must be equal to 5.57 meters per second squared.

All right. So for this problem, we have two pendulums ones on the planet Earth and ones on some unknown planet. And both have the same length string. However, they're on different planets, so they're experiencing different gravity's, and they both have separate periods. So the period of the pendulums on earth is 0.65 seconds in the period of the pendulums on the unknown, Planet P is equal to 0.862 seconds. And the first thing we have to answer is is the gravitational field on the unknown planet stronger or weaker than the gravitational field on Earth? So we're gonna look at the equation for the period of a pendulum is equal to pi times the square root of length of pendulum divide by the gravitational field of experiencing, and we're going to you analytically say All right, So if G is larger, then we're going to have a smaller period. Since she's in the denominator of this equation. If g of smaller, they won't have a larger period because in the nominee of the equation, there's an inverse law going on here. So then we're going to analyze the period of the pendulum on the Unknown Planet. We noticed that it's greater. It is greater than TCBY or he just read it like this mathematically. So if TCP is greater than TCBY, then Jesus Api. The gravity on the Unknown Planet must be lesser because there's an inverse law here. We state that the period of any pens alone is proportional to the inverse of the gravitational field with pendulum is experiencing. That's the first thing we ventured. We already know that G is going to be lesser on the Unknown Planet than it is gonna be on Earth simply due to this relation here. Now we just have to learn what is Guille neo known planet? Well, we can say we can compare the two periods and thusly compare the two gravity's through their equations. So let's do that. Let's say T sub e for the period of the pendulum Earth divided by the period of the pendulum on the Unknown Planet, it's simply going to be equal to their individual equations. So it's going to be two pi times the square root of the length of the pendulum. And remember, the length of the pendulum doesn't change that the same length pendulum on both Earth and the Unknown Planet divided by the gravity on earth. And this whole product is going to be divided by the equation for the period on the Unknown planet two pi times square, root of the length divided by the gravity on the Unknown Planet. All right, so what immediately cancels out here? The two pies cancel out so you can rewrite this and we can extend the square root to cover over this whole set of three fractions. If we do our fraction division correctly, we should notice that the else cancel this whole thing legal to the square root of geese API divided by gcb. All right, so because we have values for Tee Savi and TCP, we can plug them into this equation here and swear the whole equation today to get geese api Overdue city out of the square root. So it's plug in the values and then swear them. Let's say TV 0.65 This would be 0.65 divided by TCP, which is 0.862 We want to square this little thing and say that's equal to gee soapy, divided by gcb Now this is Multiply gcb over and evaluate this. If we evaluate this, it's going to be equal to 0.56 mine. And because we know the value of gcb to be 9.8, the gravity on Earth gravitational acceleration on earth 0.569 times 9.8 is equal to the gravitational acceleration on the planet peak. It's the gravitational acceleration. The Planet P must be equal to 5.57 meters per second squared.


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