# How to find amplitude in SHM problem

Tags:
1. Jul 1, 2016

### Brittany King

1. The problem statement, all variables and given/known data
A mass of 120g rolls down a frictionless hill reaching a speed of 4.2 m/s and collides with another mass 3.00g attached to a spring of constant 30 N/m. The two masses stick together and enter into periodic motion. What is the equation for the motion?

2. Relevant equations
m1v1+m2v2=mtV
x(t)=Acos(wt+phi)
w=k/m^1/2
Vmax=wA

3. The attempt at a solution
I found Vfinal=1.2 m/s
Also found w=8.45 rad by w=k/m^1/2
I thought A would be 0.142 but in the answer A=0.266
the answer is: x(t)=0.266sin(8.44t)

2. Jul 1, 2016

### TSny

I guess you meant to type 300 g rather than 3.00 g for the second mass.

Your work looks good to me. I agree that A = 0.142 m.

[EDIT: I'm assuming that the first mass reaches the bottom of the hill and then moves horizontally before hitting the spring as shown below.]

(If there is no friction on the hill, the 120 g mass would slide rather than roll down the hill.)

#### Attached Files:

• ###### pic.png
File size:
1.6 KB
Views:
80
Last edited: Jul 1, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted