Finding Vmax of a Block in SHM: Mass, Spring Constant, and Distance

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To find the maximum velocity (Vmax) of a block in simple harmonic motion (SHM), the block's mass is 200g, the spring constant is 5.0 N/m, and it is displaced 5.00 cm from equilibrium. The correct approach involves using the relationship between potential energy and kinetic energy, specifically applying the work-energy theorem. The calculated maximum velocity is 0.250 m/s, which aligns with the choice provided in the multiple-choice question. The discussion emphasizes the importance of correctly applying oscillation formulas and maintaining unit consistency throughout the calculations. Understanding these principles is crucial for solving similar SHM problems effectively.
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Homework Statement


A block with mass m=200g is attached to a spring with a elastic constant of k=5.0 N/m.
The block is pushed at a distance x=5.00cm of its equilibrium position, in a surface with no friction.
Then its dropped of that position. Assume for t=0s that the block is at rest.
What is the maximum velocity of the block?
3a1cb6fe8d8e320aa4a4bf6382213e13.png

This is a multiple choice question, solutions being
a) 0.250 m/s
b) 1.25 m/s
c) 0.05 m/s
d) 5.0 m/s

Homework Equations


N/A

The Attempt at a Solution


bb927676ea389d454c9ec6b4ae65d102.png

I can't figure how to find Vmax[/B]
 
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ok, that is Time Period and/or angular frequency ... do you want to use Hooke's Law (PE into KE), or do you want to use oscillation formulas?
 
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Oscillations I guess, I really can't figure out how to find Vmax. I've tried the Vmax=Aw but somehow it resulted in 1, which isn't the answer.

Vmax = Aw = (0.2 kg)(5 rad/s) = 1.

I'm really confused
 
in oscillations, capital A is the oscillation Amplitude (not the mass)
You need to keep track of your UNITS !
 
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Ok then so Vmax = Aw = (0.05)(5) = 0.250 m/s?

It makes sense now.
 
I think you can simply use the work energy theorem here.
 
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