# Simple Harmonic Motion

## 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?

## 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.

Related Introductory Physics Homework Help News on Phys.org
BvU
• 