Working out velocity of particle moving in SHM

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SUMMARY

The discussion focuses on calculating the velocity of a particle in Simple Harmonic Motion (SHM) with an amplitude of 9.3 cm, leading to an equilibrium position of 4.65 cm. The particle's displacement is determined to be 2.65 cm when it is 2 cm away from one endpoint. The calculated speed of the particle is confirmed to be approximately 0.16 m/s, validating the approach taken in the calculations.

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  • Understanding of Simple Harmonic Motion (SHM)
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  • Familiarity with displacement calculations
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Bolter
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Homework Statement
Calculate the velocity of a particle moving in SHM at a point
Relevant Equations
a = –w^2x
Screenshot 2020-02-25 at 17.46.51.png


So the way I have gone about it is to assume that the equilibrium position is half way between the 2 end points, hence the amplitude of this motion is 9.3/2 = 4.65 cm

Therefore the displacement of the particle when it is 2cm away from one end point should be the distance between that point and the equilibrium position. So i.e. 4.65 – 2 = 2.65 cm

Knowing those 2 values I have done this

IMG_3990.JPG


Is a speed of 0.16 m/s be what you should get?

Thanks for any help!
 
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Looks about right.
 
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