Can the Period of a Damped Oscillator Be Detected Accurately?

[kgal]

Homework Statement

A student measures the unstretched length of a spring as 11.2 cm. When a 100.0 g mass is hung from the end of the spring, its length is 20.7 at rest. The mass-spring system is set into oscillatory motion and the amplitude of the motion decreases to half its original value in 5 complete oscillations.
a. What is the period of the oscillatory motion, assuming no damping?
b. The student can measure the period of oscillation to an accuracy of 0.05s. Will the student be able to detect the difference between the period calculated with no damping and the period of the damped oscillator?

Homework Equations

F = -kΔx
T = 2∏/√(k/m)
Δx = x2 - x1

The Attempt at a Solution

a. Δx = x2 - x1 = 20.7 - 11.2 = 9.5 cm = 0.095 m
F = -kΔx
-k = F/Δx = mg/Δx = .98/.095 = 10.31 N/m
T = 2∏ / √(k/m) = .62 s

b. Not sure how to even start with this one...

 

[Spinnor]

I think this is the formula you need.

See,

http://farside.ph.utexas.edu/teaching/315/Waves/node9.html

https://www.google.com/webhp?hl=en#...w.,cf.osb&fp=cb785b2a153d08fa&biw=734&bih=439

 

[kgal]

is the original amplitude the difference between the spring natural length and the spring length after the mass is put on it?

 

[Spinnor]

No, the amplitude is not given and not needed for the problem. Just assume it is some value and what is important is what ever the value is it halves as stated in the problem.

 

[kampfer]

What about part b?
Did you find difference between the period calculated with no damping and the period of the damped oscillator?