How Does Acceleration Affect the Period of a Simple Pendulum?

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Homework Help Overview

The discussion revolves around the effects of acceleration on the period of a simple pendulum, specifically in different accelerating frames such as an elevator and a truck. The problem involves understanding simple harmonic motion and the relevant equations for calculating the period under varying conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the formula for the period of a pendulum and how acceleration affects it. There are attempts to derive the period using different values of effective gravity based on the acceleration of the frame. Questions arise about the implications of horizontal versus vertical acceleration on the calculations.

Discussion Status

Some participants have made progress in understanding the first part of the problem, while others are still seeking clarity on the second part involving horizontal acceleration. There is an ongoing exploration of the correct formulas and assumptions regarding effective gravity in different scenarios.

Contextual Notes

Participants note the presence of a figure that illustrates the setup for one of the problems, which may be influencing their understanding. There is also mention of a coefficient of static friction relevant to the second problem, indicating additional constraints that need to be considered.

cissablecat23
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1a)A simple pendulum is 4.35 m long. What is the period of simple harmonic motion for this pendulum if it is hanging in an elevator that is accelerating upward at 5.90 m/s2?


b)What is the period of simple harmonic motion for this pendulum if it is placed in a truck that is accelerating horizontally at 5.90 m/s2?

2) A large block P executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency of f = 1.52 Hz. Block B rests on it, as shown in the figure below, and the coefficient of static friction between the two is μs = 0.630.

What maximum amplitude of oscillation can the system have if block B is not to slip?

1)l=4.35 m
a= 5.90 m/s/s

w^2=g/l
w^2=9.80/4.35
w^2= 2.252873563

then i have to find T.. but i don't know what formula to use...

2) w=2(pie)f
w-2(pie)(1.52 Hz)
w= 9.5504
and i don't know what else to do
 
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1)

T= 2pi * sqrt(L/g)
T= 2pi* sqrt(4.35 m/ [9.8m/s^2-5.90m/s^2])

2)

"as shown in the figure below" <- picture?
 
it was a picture with a big block on the bottom.. a small block on the top.. with a spring attached to it..

i got the first part of the question.. it was T=2pi sqrt(l/g+a)

but i cannot get the second part of it... with it acclerating horizontally..
 
springs? I thought we were talking about pendulums?! I can't visualize #2
 
#2 is different from #1... it's.. a small block on a big block.. and the big block has a spring attached to it.. the coefficient of static friction was..0.630.. they want to know what's the maximum amplitude of the oscillation so that the little block doesn't slip..

f= 1.52 Hz

AND for #1.. if they change the a from upward.. to horizontal.. how does that change the answer...
 
spring attached to it where? On the bottom? sides? Need to be specific

I don't think the answer for #1 changes
 
i've got it.. thanks.. :)
 
Are you sure it's 2 pi ( sqrt ( L / (g - a ) ) ? Or is it g + a?
 
cissablecat23 said:
1a)A simple pendulum is 4.35 m long. What is the period of simple harmonic motion for this pendulum if it is hanging in an elevator that is accelerating upward at 5.90 m/s2?


b)What is the period of simple harmonic motion for this pendulum if it is placed in a truck that is accelerating horizontally at 5.90 m/s2?
In a non-accelerating frame, the period of a simple pendulum is:
[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]
In an accelerating frame, the effective "g" is different.

For case (a), the apparent acceleration is [itex]-g\hat{y} -a\hat{y} = -(g+a)\hat{y}[/itex], so [itex]g_{eff} = g + a[/itex].

For case (b), the apparent acceleration is [itex]-g\hat{y} -a\hat{x}[/itex], so [itex]g_{eff} = \sqrt{g^2 + a^2}[/itex].
 

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