- #1

Zynoakib

- 70

- 0

## Homework Statement

A particle of mass 4.00 kg is attached to a

spring with a force constant of 100 N/m. It is oscillating

on a frictionless, horizontal surface with an amplitude

of 2.00 m. A 6.00-kg object is dropped vertically on top

of the 4.00-kg object as it passes through its equilibrium

point. The two objects stick together. What

is the new amplitude of the vibrating system after the

collision?

## Homework Equations

## The Attempt at a Solution

Angular frequency before impact adding the block = (100/ 4)^1/2 = 5 s^-1

Angular frequency after impact adding the block = (100/ 10)^1/2 = 3.16 s^-1

Max acceleration before adding the block = Angular frequency^2 x amplitude = 25 x 2 = 50 ms^-2

Force of max acceleration = 50 x 4 = 200N

After addition of the block, the max acceleration will be 200N/ (6 + 4) = 20 ms^-2

Max acceleration after adding the block = Angular frequency^2 x amplitude

20 = 3.16^2 x amplitude

amplitude = 2

I know it is wrong but what is wrong, other than the fact my answer is exactly the same as the old amplitude.