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Oscillations: Simple Pendulum?
A simple pendulum of mass 19 kg is displaced an angle of 13 degrees from vertical and released. It now has a period of 2 sec.
a) Its mass is doubled. What is its period now?
T2m = 2 sec
b) Its length is doubled. What is its period now?
T2L = 2.828 s
c) What if the original pendulum is only displaced a distance of 6.5 before being released. What is its period now?
TΘ/2 = 2 sec
d) The original pendulum is taken to a planet where g = 15 m/s2. What is its period on that planet?
T = 1.6174 s
e) Let's go back to the original pendulum of mass 19 kg with a period of 2 sec, displaced an angle of 13 degrees from the vertical. What would its acceleration be in the vertical (y) direction as it reachs the lowest point on its swing?
ay = _______ m/sec2
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My only trouble is with part e. Where do I start? I'm not sure what to do! Thanks!
Thanks ABHAY, but 0 isn't the answer. I thought that already unfortunately.
2 Answers
- 1 decade agoFavorite Answer
acceleration of a simple pendulum = w^2 * x
where x is the distance traveled and w is omega angular frequency
a=0
if seen practically it is in stable equilibrium.so net force is 0
it goes up again due to its virtue of velocity not due to force
- ?Lv 44 years ago
it quite is a reasonably common problem which may be present day in a million places interior the information superhighway. seem up Pendulum action and you will locate all you want. (by utilising the way, the oscillations matter purely on gravity/length)
