What Is the Minimum Time for SHM Particle to Travel Between Two Points?

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Homework Help Overview

The problem involves a particle undergoing simple harmonic motion (S.H.M.) with a specified amplitude and period. The original poster seeks to determine the minimum time required for the particle to travel between two points, each located 12.5 cm from the mean position.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the interpretation of the distance between points and whether it refers to the distance from the mean position or the total distance between the two points. There are attempts to clarify the setup of the problem and the meaning of the time calculated.

Discussion Status

Several participants are exploring different interpretations of the problem and the calculations involved. Some have provided solutions and questioned their validity, while others are attempting to reconcile their understanding of the motion and the equations involved. There is ongoing dialogue about the correct approach to finding the time required for the particle's movement.

Contextual Notes

There are mentions of confusion regarding the use of angles in radians versus degrees, as well as the implications of negative time in the context of the problem. Participants are also considering the implications of the S.H.M. equations and how they relate to the specific distances involved.

  • #31
gracy said:
Because trig functions have always angles (either in radians or in degrees) as an argument / input . Right?
Yes.
gracy said:
What other quantities can become argument of trig functions?
As far as I know, no other.
 
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  • #32
Thank you so so much my friend @blue_leaf77 . God bless you :smile::smile:
 
  • #33
gracy said:
Thank you so so much my friend @blue_leaf77 . God bless you :smile::smile:
Glad to help you.
 
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  • #34
One more question. We can't take directly y=25 cm (12.5+12.5)cm i.e distance between points (points of interest) a & b
pointsab.png

because "y" in the equation ##y##=##a## sin ##w####t## is distance from mean position . I also tried to calculate it
25=25 sin (120 t)
1=sin (120 t)
(120 t)=90
t=90/120
t=0.75 s
Look, I got wrong answer, I should've got 0.5 s. This proves what I just wrote above. Right?
 
  • #35
gracy said:
because "y" in the equation yyy=aaa sin wwwttt is distance from mean position .
Yes, that exactly the reason.
gracy said:
25=25 sin (120 t)
1=sin (120 t)
(120 t)=90
t=90/120
t=0.75 s
You are basically calculating the time needed from the mean position to the point of farthest displacement - the amplitude. That's why what you got is a quarter of the period.
 
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