# Hooke’s Law / SHM

**1. The problem statement, all variables and given/known data**

A 1.50 kg ball and a 2.00 kg ball are glued together with the lighter

one below the heavier one. The upper ball is attached to a vertical ideal

spring of force constant 165 N/m and the system is vibrating vertically with

amplitude 15.0 cm. The glue connecting the balls is old and weak, and it

suddenly comes loose when the balls are at the lowest position in their

motion.

**Find the amplitude of the vibrations after the lower ball has come loose.**

**2. Relevant equations**

ΣF=ma

W=mg

F_{s}=-kx

m_{1}=1.5kg

m_{2}=2kg

**3. The attempt at a solution**

∑F=ma

-kx_{1}-(m_{1}+m_{2})g=0

x_{1}=(m_{1}+m_{2})g/k

x_{1}= .2079m

.15m+x_{1}=.3579m

x_{2}=(m_{2})g/k

x_{2}=.11879m

.3579m – .11879m = .239m

Even after solving it, I do not really understand what is going on in the problem. Why is x=.2079 with both of them attached, when the amplitude is .15m?

http://ift.tt/1ffXBw8

## Leave a comment