Solve Horizontal Spring Homework: T, A, V&Where Max V

[alex91alex91alex]

Homework Statement

A 425 gram mass is attached to a horizontal spring. The spring is known to have a spring constant of 84.6 N/m. The mass is free to slide along a friction less surface.

a) The mass is then pulled to the side a distance of 5.67 cm and released. Determine the period and amplitude of the resulting oscillations.

b) Determine the magnitude of the maximum velocity of the mass, and state where it reaches that maximum velocity.

Homework Equations

T=2πsqr(m/k)
ω=sqr(k/m)
KE(block) = PE(spring) =>1/2mv^2 = 1/2kA^2

The Attempt at a Solution

a) The mass is then pulled to the side a distance of 5.67 cm and released. Determine the period and amplitude of the resulting oscillations.

T=2πsqr(m/k)=2πsqr(0.425/84.6)=0.455s
ω=sqr(84.6/0.425)=14.1rad/s

Now, my question is, can i plot ω as v, in order to find the amplitude? Ain't the angular velocity suppose to reduce over time?

b) Determine the magnitude of the maximum velocity of the mass, and state where it reaches that maximum velocity.

I do not even know how to get started here, can not find any formulas.

 

[alex91alex91alex]

Oh, found out that v=rω so v=0.0567m(14.1rad/s)=0.8m/s

So,
KE(block) = PE(spring) =>1/2mv^2 = 1/2kA^2
(1/2)(0.425kg)(0.8m/s)^2=(1/2)(84.6N/m)(A^2)
A=sqr(23.0112)=4.80m

That does not seem right...

 

[alex91alex91alex]

Could really use some insight, thank you.

 

[CWatters]

The mass is then pulled to the side a distance of 5.67 cm and released.

What does the "point of release" correspond to in terms of a point in the cycle? eg What's the velocity at that point? Perhaps think about a pendulum, how you start it swinging and the amplitude that results.

PS: Figure out the amplitude before trying to calculate the velocity.

 

[CWatters]

Ain't the angular velocity suppose to reduce over time?

ω = 2πf

so if the angular velocity changed over time the frequency would also change. Does it?

 

[CWatters]

T=2πsqr(m/k)=2πsqr(0.425/84.6)=0.455s

I made it 0.445s

ω=sqr(84.6/0.425)=14.1rad/s

I agree.

Oh, found out that v=rω so v=0.0567m(14.1rad/s)=0.8m/s

That's for an object moving in a circle.

So,
KE(block) = PE(spring) =>1/2mv^2 = 1/2kA^2
(1/2)(0.425kg)(0.8m/s)^2=(1/2)(84.6N/m)(A^2)
A=sqr(23.0112)=4.80m

Figure out the amplitude first then use conservation of energy to find the velocity.