# Word problem

1. Apr 21, 2004

### OptimusPrime

An object weighing 128 pounds is suspended from a spring, streching the spring 2 feet beyond its natural length. The object is then released from rest at a point 6 inches above the equilibrium position.

a. Find a function which describes the position of the object at any time t.
b. What is the position of the object at time t= pi/12 sec? be precise
c. Is the object moving upward or downward at this time? Explain

2. Apr 21, 2004

### OptimusPrime

able to get some done but stuck

mg = kx ..... to solve for 'k'"

Okay, "mg" is the weight of the object, which we are told is 128 pounds. And "x" is the distance the spring is stretched by that weight- we are told that's 2 feet.

mg= kx is 128= 2k. What's k?

mx" + kx = 0"

Since mg= 128, m= 128/g= 128/32.2= 3.97.

k = 64

m =128/32 = 4

4x" + (64)x = 0

x" + (4^2)x = 0

x = A * cos(4t) + B * sin (4t)

Is this right? I don't know how to finish the problem. Help please

3. Apr 21, 2004

### OptimusPrime

Please can anyone help. Am struggling with this and have to hand it in first thing tomorrow. I would really really appreciate it.

Thanks

4. Apr 21, 2004

### matt grime

resolve the forces on the object using hooke's law, and then use F=ma = "mx double dot"

5. Apr 21, 2004

### OptimusPrime

I don't know how to do that can you please show me how

6. Apr 22, 2004

### jepe

The system is an undamped mass-spring system.
the natural period of the system is:
T = 2*pi*sqrt(m/k) = 2*pi*sqrt(128/64) = 2*sqrt(2)*pi
This may help you to calculate the position at any time using the equation:
x(t) = 6*cos(2*pi*t/T) (inch)
for t=0 : x(0) = 6 inch (start position)
for t=T/2 : x(T/2) = -6 inch (lowest point after half cycle)
for t=T : x(T) = 6 inch (after one full harmonic cycle, assuming no damping force)
you should be able to answer questions b and c now.