SHM question (Is my method correct?)

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Homework Help Overview

The discussion revolves around a problem involving simple harmonic motion (SHM) where a particle's motion is analyzed in terms of its speed and the time spent at certain velocity thresholds. The original poster attempts to determine the angular frequency of the particle based on the time it spends moving at speeds less than half of its maximum speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the symmetry of SHM and the time intervals associated with different speed thresholds. There are attempts to derive the angular frequency using trigonometric relationships and interpretations of the motion graphically.

Discussion Status

Some participants have provided guidance on visualizing the problem through drawings, while others question the original poster's method and reasoning. Multiple interpretations of the time spent at various speeds are being explored, and there is an ongoing dialogue about the correctness of the calculations presented.

Contextual Notes

There is mention of a specific time duration (4/3 seconds) and its relation to the full period of the motion, which is suggested to be 4 seconds. The discussion includes references to the fractions of the period corresponding to different speed conditions.

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Homework Statement


A particle P is performing simple harmonic motion with amplitude 0.25m. During each complete
oscillation, P moves with a speed that is less than or equal to half of its maximum speed for 4/3 seconds.
Find the angular frequency of P

The Attempt at a Solution


First I split 4/3 into two portions, the right and left side. Since both are symmetrical, information about left side will be same as information for right side.

Now 2/3s is the time the particle spend with velocity < 1/2 wx in one side of motion.
So
Time Spent = Time required to achieve max - time required to achieve 1/2 of max
$$ \frac{2}{3} = \omega^{-1} (\sin^{-1}(\frac{Max~V}{Max~V})-\sin^{-1}(\frac{1/2~of~Max~V}{Max~V})) $$
$$ \frac{2}{3} = \omega^{-1} (\sin^{-1} 1 - \sin^{-1} 0.5) $$
$$ \omega = \frac{\pi}{2} $$
Note:- The answer is correct

Now the problem I realized it, even though my answer is coming correct but in my opinion, my method is not correct. The reason is I have calculated the time required by the particle to go from centre to the 1/2 speed points which makes no sense.
So can you tell me whether my method is correct and if not what is the correct method?
 
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Hi,

If you draw a sine and check what fraction of the period the absolute value is < 0.5 the answer is 1/3 .
So 1/3 of a period corresponds to 4/3 s. Thereby : a full period is 4 s, hence ##\omega = \pi/2## as you 'found'.
Faiq said:
Now 2/3s is the time the particle spend with velocity < 1/2 wx in one side of motion.
So
Time Spent = Time required to achieve max - time required to achieve 1/2 of max
Don't agree: make a drawing.
 
BvU said:
Hi,

If you draw a sine and check what fraction of the period the absolute value is < 0.5 the answer is 1/3 .
So 1/3 of a period corresponds to 4/3 s. Thereby : a full period is 4 s, hence ##\omega = \pi/2## as you 'found'.
Don't agree: make a drawing.
How did you work out the value 1/3?
 
I looked at the drawing I made :smile:
 
Î made a drawing but how did you infer the value fo 1/3?
 
## [0,\pi/6] + [5\pi/6, 7\pi/6] + [11\pi/6, 2\pi] ## totals ##4\pi/6##. And ##4\pi/6 /(2\pi) = 1/3##

upload_2017-2-13_17-26-40.png
 
Oh okay thanks
 
You're welcome
 

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