What is the Maximum Acceleration of a Mass Attached to a Vertical Spring?

[adashiu]

Homework Statement

A 0,824kg mass attached to a vertical spring of a force constant 162N/m oscillates with a maximum speed of 0,372 m/s. Calculate the maximum acceleration of the mass.

The Attempt at a Solution

a_max=v_max*\omega

\omega=\sqrt{\frac{k}{m}}

Is that correct?

 

[Doc Al]

Yes, that will work.

 

[Li109]

I am a bit confused on how to calculate the acceleration of the mass. Are the accelerations at the bottom and the top the same or different? at eq we have kx=mg. when you displace the mass downward and release, ma=kx-mg. when the mass reaches the top, kx+mg=ma. since x is the same in both cases, the two a's need to be different. what is the flaw in my line of thinking?

 

[Doc Al]

I am a bit confused on how to calculate the acceleration of the mass. Are the accelerations at the bottom and the top the same or different? at eq we have kx=mg. when you displace the mass downward and release, ma=kx-mg. when the mass reaches the top, kx+mg=ma. since x is the same in both cases, the two a's need to be different. what is the flaw in my line of thinking?

Realize that the mass oscillates about the equilibrium point, where kx₀ = mg.

At the lowest point, the net force is: k(x₀ + A) - mg = kA (upward)
At the highest point, the net force is: k(x₀ - A) - mg = -kA (downward)

 

[Li109]

Realize that the mass oscillates about the equilibrium point, where kx₀ = mg.

At the lowest point, the net force is: k(x₀ + A) - mg = kA (upward)
At the highest point, the net force is: k(x₀ - A) - mg = -kA (downward)

i still don't understand. can u please explain further?

 

[Li109]

so this means that the force due to the spring at the top and the bottom are different? while if the system was horizontal, it'd be the same at both extremes?

 

[Doc Al]

so this means that the force due to the spring at the top and the bottom are different?

Right.

while if the system was horizontal, it'd be the same at both extremes?

If the system was horizontal, the equilibrium point would be at the unstretched position, x = 0. Thus at one extreme the spring force would be kA and at the other it would be -kA.