Why Do Different Methods Yield Different Results for Simple Harmonic Motion?

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Homework Help Overview

The discussion revolves around a horizontal spring-mass system on a frictionless table, exploring the calculations of time taken to complete a quarter of a cycle in simple harmonic motion using different methods. Participants are examining the implications of using average acceleration versus conventional methods based on the spring constant and mass.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss two methods for calculating the time for a quarter cycle, one using conventional formulas and the other involving average acceleration. Questions arise regarding the validity of assuming average acceleration and the implications of non-uniform acceleration in the context of simple harmonic motion.

Discussion Status

The discussion is active, with participants questioning assumptions about average acceleration and its applicability to the spring-mass system. Some participants have provided insights into the nature of acceleration in this context, noting that it is not linear with respect to time but rather sinusoidal with respect to position.

Contextual Notes

There is an ongoing debate about the assumptions made in the calculations, particularly regarding the treatment of average acceleration and its relationship to the motion of the spring. Participants are also considering the implications of these assumptions on the results obtained from different methods.

Zolo
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There is a horizontal spring-mass system lie on top of a frictionless table.
let m=mass
k=spring constant
x=displacement from equilibrium position
A=amplitude
T=period

By using conventional way, time taken to complete a quarter of cycle=T/4=(∏/2)√(m/k)

consider another way of doing, F=-kx
a=-(k/m)x
then we can consider the average acceleration of first quarter of motion a=(1/2)(k/m)A
since s=ut+1/2at^2
then, A=(1/2)*(1/2)(k/m)A*t^2
finally, we get t=2√(m/k)

why is the ans from both way diffrent...Both ways seem to be equivalent...
 
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Are you simply assuming that's the average acceleration? ... or can you show it?

The acceleration due to the spring, is not uniform.
 
If u consider a body moving with constant acceleration(constant rate of change of velocity with time), then we know that average velocity across any time interval=(V2-V1)/2
Thus, the motion can be regarded as constant velocity with the average velocity

In this system, rate of change of acceleration with respect to x is constant, so when can conclude that average acceleration across any interval of x=(a2-a1)/2
For this case,
a2= acceleration at maximum displacement=(k/m)A
a1=acceleration at equilibrium position=0

isn't this show that the motion can be regarded as a constant acceleration with the average acceleration?
 
Its true that if velocity is linear with respect to time, then the average velocity can be treated as a constant velocity. Its also true that if acceleration is linear with respect to time, then the average acceleration can be treated as a constant acceleration. But in the case of the spring, the acceleration is not linear with respect to time, its linear with respect to position. Your method would be fine if acceleration were linearly related to time, but its not; its sinusoidaly related.
 
Last edited:
In fact, over one complete cycle of the harmonic oscillator, the average acceleration is zero !
 
Its true that if velocity is linear with respect to time, then the average velocity can be treated as a constant velocity. Its also true that if acceleration is linear with respect to time, then the average acceleration can be treated as a constant acceleration. But in the case of the spring, the acceleration is not linear with respect to time, its linear with respect to position. Your method would be fine if acceleration were linearly related to time, but its not; its sinusoidaly related.

Can you explain or show why it so when it is linearly related to time?
 

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