# What is the period of oscillation of the mass on the spring?

## Homework Statement

a)When a mass is attached to the end of a vertical spring, the spring is stretched down 3 cm. If the mass is pulled down a bit farther and then released, what is the period of oscillation of the mass on the spring? b) if the mass moves through its equilibrium position at 50 cm/s, what is the amplitude of the oscillation?

## Homework Equations

T= 2pi/v , w= srqt(g/L) , T=2pi/w

## The Attempt at a Solution

I tried using w= srqt(g/L) because the problem only gives you with the radius of the ball, so i just that equation to find the angular velocity and than use w to find the period with T=2pi/w but i don;t get the right answer. which is 0.348 seconds.
for part b i just don;t even know how to start.
I need help?

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gneill
Mentor

## Homework Statement

a)When a mass is attached to the end of a vertical spring, the spring is stretched down 3 cm. If the mass is pulled down a bit farther and then released, what is the period of oscillation of the mass on the spring? b) if the mass moves through its equilibrium position at 50 cm/s, what is the amplitude of the oscillation?

## Homework Equations

T= 2pi/v , w= srqt(g/L) , T=2pi/w

## The Attempt at a Solution

I tried using w= srqt(g/L) because the problem only gives you with the radius of the ball, so i just that equation to find the angular velocity and than use w to find the period with T=2pi/w but i don;t get the right answer. which is 0.348 seconds.
for part b i just don;t even know how to start.
I need help?
Non sequitur. What ball? Why angular velocity when you're dealing with a linear spring?

tms
What radius of what ball? This is just a mass on a spring.

You want to look at the kinetic and potential energy. Write down the equation for energy at the beginning, and write down the equation for energy at the equilibrium position.

Got it!!!! yeah i really didn't have to use any angular velocities. The hard part was assuming that the mass is 1 kg. because it only mentions mass and they don;t mention anything else so i tried 1 kg just for curiosity.

I used T=2pi*sqrt(m/k) i knew that k=F/deltay so (1kg)(9.8)/0.03m=326.7 N/m than plug everything in T=2pi*sqrt(m/k) and i got 0.347seconds which was the right answer. Thank gneill!

than for the second part I used A=VmaxT/2pi and got the right amplitude too.

NascentOxygen
Staff Emeritus