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 2 years ago
3. An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion, as shown in this graph.
How many seconds out of phase with the displacements shown would graphs of the velocity and acceleration be?
 2 years ago
3. An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion, as shown in this graph. How many seconds out of phase with the displacements shown would graphs of the velocity and acceleration be?

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whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1The velocity graph: velocity is the rate of change of displacement, so velocity is going to be 0 when the pendulum changes direction at the end of each swing, and nonzero when it is moving. The acceleration graph: acceleration is the rate of change of velocity, so it will be 0 as the pendulum swings through the lowest point (= midpoint) of its swing. Looks like 9 complete swings take 5.0 seconds, so 1 swing is about 0.56 s. I think the velocity graph is 1/4 of a full cycle behind the displacement graph, about 0.14 s. The acceleration graph has the same general shape as the displacement graph, except it is inverted and multiplied by a scale factor which is proportional to the force of gravity and inversely proportional to the length of the pendulum. That means it is half a cycle out of phase, about 0.28 seconds. I'm not sure why the length of the pendulum is given, perhaps to confuse us?

erin512
 2 years ago
Best ResponseYou've already chosen the best response.0@whpalmer4 thank god for you. nobody else on here knows how to do this stuff. thanks so much for the explanation as well. is 0.28 the answer?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1There are two answers, 0.14 second for the velocity curve and 0.28 second for the acceleration curve. Remember, the angular acceleration for small displacements is (g/R)sin theta, and the angular velocity is the integral of the acceleration, so it is going to be a cos theta expression. This whole answer may well be worth exactly what you've paid for it :)

erin512
 2 years ago
Best ResponseYou've already chosen the best response.0@whpalmer4 thank you so much :) my deadline for all of this is today, i have a big question that i have nooo earthly idea how to solve, if i post it do you think you could help me with it?
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