A mass of 1.32 kg is connected to a spring of spring constant 8.81 N/m ....
A mass of 1.32 kg is connected to a spring of spring constant 8.81 N/m . An oscillation is started by pulling the mass to the right to amplitude 0.582m before release and the oscillator moves in air. The oscillation decays to 18.2% of the original amplitude in 58.2 seconds.
the damping constant of the oscillation is 7.73*10^-2 kg/s
total energy has the system lost in this time due to air damping = 1.44 j
the amplitude of the oscillation be 25.29 seconds after release is 0.278
What would the position of the oscillation be 25.29 seconds after release?
Homework Answers
The position of the oscillator after 25.29 sec after release is
x(t) = A cos (wt + )
A = amplitude after 25.29 sec = 0.278
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A mass of 1.32 kg is connected to a spring of spring constant
8.81 N/m ....
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