How to find amplitude in SHM problem

  • #1

Homework Statement


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?

Homework Equations


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

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)
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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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.)
 

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