Simple Harmonic Motion, Initial Displacement vs Initial Cond

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SUMMARY

The discussion centers on the nuances of simple harmonic motion (SHM), specifically regarding initial displacement and conditions. The initial displacement of 6 inches is referenced, but the author uses -1/2 as the initial condition at t=0, indicating a fractional representation of peak-to-peak amplitude. The negative sign reflects the direction of the restoring force, opposing the displacement. Additionally, the amplitude calculated later is larger than the initial displacement due to initial velocity, confirming that the object does not start from rest.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with displacement and amplitude concepts
  • Knowledge of restoring forces in oscillatory systems
  • Basic grasp of units of measurement, particularly imperial units
NEXT STEPS
  • Study the mathematical equations governing simple harmonic motion
  • Learn about the relationship between initial displacement and amplitude in SHM
  • Explore the effects of initial velocity on oscillatory motion
  • Investigate the use of different units in physics, focusing on imperial vs metric
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Students and educators in physics, engineers working with oscillatory systems, and anyone interested in the mathematical foundations of simple harmonic motion.

austrosam
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Hi everybody,

I'm writing an exploration on the mathematics of simple harmonic motion and I stumbled across something I fail to understand in one of my resources (http://tutorial.math.lamar.edu/Classes/DE/Vibrations.aspx). In the example the author uses toward the end of the resource, the object is initially displaced by 6 inches (don't ask me why he felt the need to use imperial units) but then, the initial condition for displacement given at t=0 is -1/2. Should it not be 6?

My guess is that one can simply set t=0 at any point during the oscillation and not in fact when the oscillation is started, but that still would not quite explain everything. Maybe I am just being very silly...

Many thanks for any advice!

Sam
 
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If you pull a pendulum to one side by 6" and let it go it will swing back and forth a total distance of 12". So the initial displacement is half the peak to peak amplitude. It sounds like they decided to use "fraction of peak to peak amplitude" as the unit of displacement rather than inches or meters.

The minus sign is probably because the restoring force is in the opposite direction to the displacement.
 
CWatters said:
If you pull a pendulum to one side by 6" and let it go it will swing back and forth a total distance of 12". So the initial displacement is half the peak to peak amplitude. It sounds like they decided to use "fraction of peak to peak amplitude" as the unit of displacement rather than inches or meters.

The minus sign is probably because the restoring force is in the opposite direction to the displacement.
Right, I guess that makes sense, though I must say it still seems a little odd, to me it would seem much more straightforward to use a value of 6 inches instead.

Many thanks anyway!

One more thing, I merely need quick confirmation I'm on the right track here. Later on, they calculated the amplitude which was slightly larger than the initial displacement. Is this because of the initial velocity, and the object not starting from rest?
 
I f the author is using imperial units then the lengths would be in feet - so 6 inches is 1/2 a foot. Then they would use g=32 ft/sec/sec.
 
bhillyard said:
I f the author is using imperial units then the lengths would be in feet - so 6 inches is 1/2 a foot. Then they would use g=32 ft/sec/sec.

Perfect! Thank you! I am totally unfamiliar with imperial units, I should have really checked that. Thanks!
 

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