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.
Litdaze

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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