What is the formula for calculating the maximum speed of a mass on a spring?

[MattsVai]

Homework Statement

Prove that the maximum speed (Vmax) of a mass on a spring is given by 2(Pi)fA

f = period
A = Amplitude (compression)

Homework Equations

Ek = 1/2mv^2

Et = Ek + Ee

The Attempt at a Solution

So from what I understand, Velocity is maximum when total energy is equal to kinetic energy at the point of equilibrium... this confuses me because if velocity is max at point x=0 then A would also=0 and therefore mess up the question... I am missing something... any insight is appreciated.

 

[chanvincent]

The A here stands for the Maximum Amplitude in one period, it should be a constant, regardless of the value of x, v etc...

 

[radou]

So from what I understand, Velocity is maximum when total energy is equal to kinetic energy at the point of equilibrium... this confuses me because if velocity is max at point x=0 then A would also=0 and therefore mess up the question... I am missing something... any insight is appreciated.

Since energy is conserved, the sum of kinetic and potential energy is constant. The equation of displacement of the mass is y(t) = Asin(wt). The other reltaion you need is f = w/(2Pi). Now form the potential energy, and minimize it to find the time t0 at which kinetic energy (i.e. the velocity) reaches a maximum, and plug it into the equation of velocity of the mass.

 

[chaoseverlasting]

Differentiate the equation to get expression for velocity, put it in exp for KE and maximise.

 

[MattsVai]

Thank you for the response guys but we haven't given displacement of mass yet... I am not understanding your suggestions. I am just looking for some help regarding the underlying logic... maybe from there I can infere the equations necessary to get the answer... thanks again.

 

[chaoseverlasting]

The mass on the spring undergoes SHM. So when its displaced initially by a distance x, all the energy is in the form of potential energy stored by the spring.
At this point, the spring want to push the mass outwards. PE of spring is $$\frac{1}{2} kx^2$$ where k is the spring constant and x is the displacement from the mean position.

When let go, the block accelerates away from the spring and the PE is converted into KE, which is maximum at the mean position.

When the block goes beyond the mean position, then the force applied on the block by the spring acts towards the mean position and the block starts retarding. Again at some point away from the mean position, the KE of the mass is converted completely into the PE of the spring.

The displacement at which the mass stops is the amplitude. Since the block starts retarding as soon as it passes the mean position, at the mean position, the KE must be maximum.

You can also derive the same result if you take the equation of displacement of SHM.

 

[MattsVai]

thanks for the explanation chaos :)

So basically, correct me if I am mistaken, but is this "mean" the point of equilibrium where x = 0? Also, would this imply that at that precise moment elastic energy is zero?

If so then Et = Ek,

Et = 1/2mv^2 right?

Where does frequency fit in... humm

 

[MattsVai]

Figured it out... thanks guys.