Homework Help: Simple harmonic motion, spring, block on block, static friction help

1. May 7, 2008

scholio

been working on this for a long time, havent got far, keep getting a negative mass!

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

A 440 g block on a frictionless surface is attached to a rather limp spring of constant k = 8.7 N/m. A second block rests on the first, and the whole system executes simple harmonic motion with a period of 1.4 s. When the amplitude of the motion is increased to 35 cm, the upper block just begins to slip. what is the coefficient of static friction between the blocks?
*mass of block on top is not given

2. Relevant equations

see below

3. The attempt at a solution

this is what i've done thus far

m1 = 440g = 0.44kg
m2 = ?
k = 8.7n/m
T(period) = 1.4s
A(amplitude) = 35cm = 0.35m
omega = angular velocity

using T = 2pi/omega --> gives omega = 4.48 m/s
using omega = sqrt(k/m) ---> where m = m1 + m2 = 0.44kg + m2 ---> solve for m2 = -0.0065 kg

that is impossible, negative mass? what am i doing wrong? one i get m2 i'll be able to figure mu static by comparing static friction force to force = ma

thanks

2. May 7, 2008

scholio

EDIT - another quick question, since the mass of the block on top was not given in the problem, does that imply that that mass is not needed? i think you do need it to find the normal force to find the force of static friction

3. May 7, 2008

DavidWhitbeck

Newton's 2nd Law on top block:

$$m_1 a = f$$

$$m_1 a =\mu m_1 g$$

$$a=\mu g$$

SHO equation for both blocks (the same acceleration because no slipping-- they accelerate together)
$$a=\omega^2 A$$

Put them together--
$$\mu=\frac{\omega^2 A}{g}$$

$$\mu = \frac{(2\pi)^2 A}{T^2 g}$$

So yeah it looks like you don't need to find the other mass.

4. May 7, 2008

scholio

thanks so much, i'll have a closer look, the answer was correct mu static = 0.72, how did you determine that you didn't need to know the mass of the block on top?

5. May 8, 2008

DavidWhitbeck

Well I didn't know a priori, I just started typing and saw that the final formula is independent of mass.