Asked by Emily
Two different simple harmonic oscillators have the same natural frequency (f=5.60 Hz) when they are on the surface of the Earth. The first oscillator is a pendulum, the second is a vertical spring and mass. If both systems are moved to the surface of the moon (g=1.67 m/s2), what is the new frequency of the pendulum?
Answers
Answered by
drwls
Frequency of a pendulum is proportional to sqrt(g/L). L stays the same on the moon, so f must be less by a factor
sqrt(1.67/9.80)= 0.412.
That would make the pendulum frequncy on the moon 0.412*5.60 = 2.3 Hz
sqrt(1.67/9.80)= 0.412.
That would make the pendulum frequncy on the moon 0.412*5.60 = 2.3 Hz
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