Pendulum Frequency Homework: 2.30 Hz

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Homework Help Overview

The problem involves a simple pendulum oscillating with a frequency of 2.30 Hz, and it explores how this frequency changes when the pendulum accelerates upward or downward at 0.500 g. The discussion centers around the effects of changing net acceleration on the frequency of the pendulum's oscillation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss whether the frequency remains constant under different acceleration conditions and explore the implications of changing net acceleration on the formula for frequency. There is also consideration of how to determine the length of the pendulum based on the given frequency and gravitational acceleration.

Discussion Status

Participants are actively engaging with the problem, questioning the assumptions about frequency constancy and discussing how to adjust the gravitational acceleration in the frequency formula. Some guidance has been provided regarding the calculation of the pendulum length and the appropriate adjustments to the acceleration values for each scenario.

Contextual Notes

There is an underlying assumption that the length of the pendulum remains constant despite the changes in acceleration. Participants are also working within the constraints of the given frequency and gravitational acceleration values.

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Homework Statement



A simple pendulum oscillates with frequency 2.30 Hz.
(a) What is its frequency if it accelerates at 0.500 g upward?
(b) What is its frequency if it accelerates at 0.500 g downward?

Homework Equations



f=1/[2(pi)(L/g)^.5]


The Attempt at a Solution



Won't frequency be constant? So both a and b are 2.30Hz?
 
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If the pendulum is accelerating upward or downward the net acceleration the pendulum is experiencing will change. So, in this case, where g normally is in the formula, you must replace with the new net acceleration, call it a. So, what would the net acceleration be for each case. If you can find these values, then you can find the frequencies.
 
G01 said:
If the pendulum is accelerating upward or downward the net acceleration the pendulum is experiencing will change. So, in this case, where g normally is in the formula, you must replace with the new net acceleration, call it a. So, what would the net acceleration be for each case. If you can find these values, then you can find the frequencies.

So how about finding L? Before I just use f=2.3Hz and g=9.81 and solved for L. Is that correct?
 
Yes, The value of L found for when the pendulum is not accelerating will be the same in parts A and B. Nothing suggests the length of the string changes.
 
G01 said:
If the pendulum is accelerating upward or downward the net acceleration the pendulum is experiencing will change. So, in this case, where g normally is in the formula, you must replace with the new net acceleration, call it a. So, what would the net acceleration be for each case. If you can find these values, then you can find the frequencies.

So for part a g=9.81-0.5*9.81
and b would be g=9.81+0.5*9.81?

Thanks
 
That looks correct.
 

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