MEDALS AWARDED. (attachment) An astronaut lands on an alien planet. He places a pendulum (L = 0.300 m) on the surface and sets it in simple harmonic motion, as shown in this graph.

Answer the following questions:
a. What is the period and frequency of the pendulum’s motion?
b. How many seconds out of phase with the displacements shown would graphs of the velocity and acceleration be?
c. What is the acceleration due to gravity on the surface of the planet in m/s2? Determine the number of g-forces.

Question

MEDALS AWARDED. (attachment) An astronaut lands on an alien planet. He places a pendulum (L = 0.300 m) on the surface and sets it in simple harmonic motion, as shown in this graph.


Answer the following questions:
a. What is the period and frequency of the pendulum’s motion?
b. How many seconds out of phase with the displacements shown would graphs of the velocity and acceleration be?
c. What is the acceleration due to gravity on the surface of the planet in m/s2? Determine the number of g-forces.

anonymous · Answer

http://assets.openstudy.com/updates/attachments/52b213dfe4b02932bbbda87a-talyorz-1387402267902-1.jpg http://assets.openstudy.com/updates/attachments/52b213dfe4b02932bbbda87a-talyorz-1387402267609-2.jpg

anonymous · Answer

From the graph, you can determine Time period by estimating how long it takes to complete one time period. Peak to Peak, trough to trough of the wave. I estimate it to be 0.75s. Frequency is just 

$$f=\frac{ 1 }{ T }$$ Expressed in Hz

In simple harmonic motion, velocity is phase shifted by 90 degrees relative to displacement, so take 1/4 of a cycle (1/4 T) as your phase difference for velocity. Acceleration is 180 degrees out of phase with displacement, so take 1/2 T as the phase difference.

In order to determine g, we use the formula for time period on a pendulum:

$$T=2\pi \sqrt{\frac{ l }{ g }}$$

rearrange to find g:

$$g=\frac{ 4\pi ^{2} L }{^{T2} }$$

You should find it's about 1/6th of Earth's acceleration due to gravity.

Hope this helps

anonymous · Answer

For the first part, would the frequency be 1 over .75?? @Dan.Harrison

anonymous · Answer

@whpalmer4 I saw you answered one similar to this.. Could you maybe help me out? I don't know if the same concepts apply since the length is different.

anonymous · Answer

@AdrDiazEs Do you know about harmonic motion, and pendulums?

anonymous · Answer

I believe the frequency would be 1/0.75 yes. But that's for the first link you sent.

anonymous · Answer

Oh okay. Thank you. The problem had both pictures underneath it.. so I'm not quite sure. I'll just use the first one. I really appreciate your help! Thank you! :)

anonymous · Answer

It's no problem!