Pendulum SHO but with extra downward acceleration of the pivot

In summary, the conversation discusses the intuitive understanding of a problem involving a pendulum in an accelerating elevator. The concept of acceleration relative to the lift is mentioned, as well as the effect this has on the period of the pendulum. The conversation also brings up the question of why a pendulum would swing in free fall and provides links to further resources for understanding. Ultimately, the conversation highlights the importance of considering all acceleration vectors when analyzing the motion of an object.
  • #1
simphys
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Homework Statement
A pendulum is hanging fromt he ceiling of an elevator. Its period (at small angles) is T sec when the elevator is at rest. We now accelerate the elevator downward w/ 5m/s^2. What is the period now? Be quantitative. [g = 10m/s^2]
Relevant Equations
##T = 2*/pi * sqrt(l / g)##
Hey guys,
Can someone help me understand how to understand this problem intuitively please?
How I understand is that I need to look the acceleration relative to the lift as if it were f.e. on another planet with a different acceleration. this gives me a = g - 5.
But then again if I didn't look up the solution I would not have been able to solve it. So.. I don't really understand this intuitively. I actually thought (before looking at the solution) that it stays the same period T as it is dependent on the gravitational acceleration.
This kind of confuses me, and leaves me feeling that I don't really understand what what the g means in the equation.Thanks in advance.
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  • #2
Think about the extreme fall case, free fall. Why would the pendulum swing at all?
 
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  • #3
simphys said:
... I don't really understand this intuitively. I actually thought (before looking at the solution) that it stays the same period T as it is dependent on the gravitational acceleration.
This kind of confuses me, and leaves me feeling that I don't really understand what the g means in the equation.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html#c3

For a simple common pendulum, gravity is the only acceleration that gets combined with its mass to produce a restoring force.
Our elevator introduces another acceleration, which vector is aligned with the gravity acceleration vector.

Therefore, a summation of those vectors would result in an increased or decreased net acceleration acting on the center of mass of our pendulum.

If the elevator is accelerating the pendulum upwards, it should be "feeling" heavier, just like you do when riding one of those, and vice-verse.

In one case, both acceleration vectors point in the same direction (vectors addition applies).
In the other case, they point in opposite directions (vectors subtraction applies).
 
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Related to Pendulum SHO but with extra downward acceleration of the pivot

What is a "Pendulum SHO but with extra downward acceleration of the pivot"?

This refers to a simple harmonic oscillator (SHO) pendulum where the pivot point is subjected to an additional downward acceleration. This means the pivot is not stationary but accelerating downward at a constant rate, modifying the dynamics of the pendulum's motion.

How does the extra downward acceleration of the pivot affect the pendulum's motion?

The extra downward acceleration of the pivot effectively increases the gravitational acceleration experienced by the pendulum. This results in an increased effective gravitational force, which can alter the period and frequency of the pendulum's oscillations.

What is the new effective gravitational acceleration?

The new effective gravitational acceleration \( g_{\text{eff}} \) is the sum of the standard gravitational acceleration \( g \) and the additional downward acceleration \( a \). Mathematically, it is given by \( g_{\text{eff}} = g + a \).

How do you calculate the new period of the pendulum?

The period \( T \) of a simple pendulum with the effective gravitational acceleration \( g_{\text{eff}} \) is given by \( T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} \), where \( L \) is the length of the pendulum. With the extra downward acceleration, the period becomes \( T = 2\pi \sqrt{\frac{L}{g + a}} \).

Does the amplitude of the pendulum's oscillation change with the extra downward acceleration?

The amplitude of the pendulum's oscillation is primarily determined by the initial conditions, such as the initial displacement and velocity. The extra downward acceleration does not directly affect the amplitude, but it does influence the period and frequency of the oscillations.

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