Two Oscillator Frequencies on the moon

AI Thread Summary
The discussion focuses on calculating the new frequencies of two oscillators, a vertical spring and a pendulum, when moved from Earth to the Moon. The frequency of the vertical spring remains unchanged despite the lower gravitational acceleration on the Moon, as it is independent of g. In contrast, the frequency of the pendulum is directly proportional to the square root of gravitational acceleration, leading to a different frequency on the Moon. Participants share methods for calculating the new frequencies, emphasizing the importance of understanding how frequency relates to gravity. Ultimately, the original poster resolves the problem independently after receiving guidance.
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Two different simple harmonic oscillators have the same natural frequency (f=2.80 Hz) when they are on the surface of the Earth. The first oscillator is a vertical spring and mass, the second is a pendulum. If both systems are moved to the surface of the moon (g=1.67 m/s2), what is the new frequency of the vertical spring and mass?

Calculate the new frequency of the pendulum. ?



This is the question. For it i just used the frequency to find the period, i then used this to find the length of the pendulum, i think solved with the new g, to find the period, and then solved back to find the new frequency, i tried this several times, and i cannot seem to get to the new frequency. I have tried similar things with the spring and mass, but i cannot find the right equations to find it.
 
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How, exactly, do the frequencies of the two oscillators depend on g? If you know that then the problem reduces to ratios and proportions.
 
Yes but it's not giving me the right answer, is there another way to go about this problem?
 
For the spring and mass, the frequency is independent of the value of g. So, it remains the same.

And for the pendulum, instead of doing this long calculation, just see that the frequency is directly proportional to the square root of the value of g.
So,

\frac{f_m}{f_e}=\sqrt{\frac{g_m}{g_e}}

Hope that helps!

spacetime
www.geocities.com/physics_all
 
thanks, it would have helped, but i finally figured it out myself :) you are correct sir in your help
 
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