Working out velocity of particle moving in SHM

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The discussion focuses on calculating the velocity of a particle in simple harmonic motion (SHM) by determining its amplitude and displacement. The amplitude is established as 4.65 cm, derived from halving the total distance of 9.3 cm. The displacement when the particle is 2 cm from one endpoint is calculated to be 2.65 cm. The participant seeks confirmation on whether a calculated speed of 0.16 m/s is accurate. Overall, the calculations and assumptions about SHM are discussed, with a positive response regarding the speed value.
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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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