# Damped Harmonic Oscillator

1. Dec 6, 2004

Hi,

I'm having a lot of trouble with a damped harmonic oscillator problem:

A damped harmonic oscillator consists of a block (m=2.00kg), a spring (k=10 N/m), and a damping force (F=-bv). Initially it oscillates with an amplitude of 25.0cm. Because of the damping force, the amplitude falls to 3/4 of this initial value at the completion of four osciallations. (a) What is the value of b? (b) How much energy has been "lost" during these four oscillations?

The truth is. I'm not really sure where to start. My book only gives a few equations to work with and I'm not sure how to relate them to find the value of b.

The first thing I did was find the period (T), by the equation
T=2∏ √(2.0Kg/10N/m) = 2.8s

...,but I'm not really sure what to do with it.

Any help with this would be awesome!

Thanks

2. Dec 6, 2004

### dextercioby

The equation for the damped oscillator along let's say the "x" axis reads:
$$m\frac{d^2 x}{dt^2}+v\frac{dv}{dt}+kx=0.$$.
The period is $T=2\pi\sqrt{\frac{m}{k}}$.The amplitude of the oscillations descreases with time exponentially as it would be shown by solving the differential equation above:
$$A(t)=A_{0} \exp({-\frac{b}{2m}t})$$.
Use tha fac that 4 periods mean a certain amount of time (4T=t) and the fact that $A(4T)=3/4 A$,plug it in the equation,simplify through A,take the logaritm,substitute t=4T and the value of m to find your answer.
Use the total energy formula (kinetic+potential)in which u plug the correct figures u have.The result should be pretty simple.The energy lost is just the difference between the one at the intial time and the one after 4T.BTW,your first calculation for the period of oscillation was correct.

Good luck!!

Last edited: Dec 6, 2004
3. Dec 6, 2004

Thanks!

Thanks,

You are awesome!

4. Dec 6, 2004

Thank You!!