- #1
jmm5872
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Suppose one pendulum consists of a 1-meter string with a bob that is an aluminum sphere 2 inches in diameter. A second pendulum consists of a 1-meter string with a bob that is a brass sphere 2 inches in diameter. The two pedulums are set into oscillation at the same time and with the same amplitude A. After 5 minutes of undisturbed oscillation, the aluminum pendulum is oscillating with one half of its initial amplitude. What is the oscillation amplitude of the brass pendulum? Assume that the friction is due to the relative velocity of bob and air and that the instantaneous rate of energy loss is proportional to the square of the velocity of the bob.
The answer is supposed to be .81A.
y(t) = Ae^(t/2[tex]\tau[/tex])cos([tex]\omega[/tex]t)
I think I might just be confused about the wording of this problem. I know that the angular frequency of a pendulum only depends on gravity and the length of the string, therefore both pendulums will be exactly the same. So now I tried to plug in 300 seconds to the above equation for t, and set A to .5A. Now I am stuck, I don't know how to get [tex]\tau[/tex].
Also, this is where my confusion comes in, if after 5 minutes the aluminum bob is oscillating at half its initial amplitude, wouldn't the brass bob be the exact same? Then the answer would be .5A.
The answer is supposed to be .81A.
y(t) = Ae^(t/2[tex]\tau[/tex])cos([tex]\omega[/tex]t)
I think I might just be confused about the wording of this problem. I know that the angular frequency of a pendulum only depends on gravity and the length of the string, therefore both pendulums will be exactly the same. So now I tried to plug in 300 seconds to the above equation for t, and set A to .5A. Now I am stuck, I don't know how to get [tex]\tau[/tex].
Also, this is where my confusion comes in, if after 5 minutes the aluminum bob is oscillating at half its initial amplitude, wouldn't the brass bob be the exact same? Then the answer would be .5A.