Calculate b/2m for Damped Oscillations of 1.00 m Pendulum at 18.0°

[nemzy]

A pendulum of length 1.00 m is released from an initial angle of 18.0°. After 500 s, its amplitude is reduced by friction to 5.5°. What is the value of b/2m?

i have no idea how to do this prooblem, the book goes over this section really briefly...

what the heck is b/2m?

 

[HallsofIvy]

Go back and read that brief section again. In particular read the problem itself carefully so you can tell us what "b" means in terms of this particular problem (I'm willing to guess that "m" is the mass of the pendulum).

 

[nemzy]

b is related to the strength of the resistance force, and the restoring force exerted on the system is -kx


they give this formula to find the angular frequency:

w= square root of [(k/m)-(b/2m)^2]

 

[PICsmith]

w= square root of [(k/m)-(b/2m)^2]

so,
b/2m = damping parameter = square root of [(k/m)-w^2],
where w = (2*pi)/(2*T),
and 2*T is the "period" of the damped oscillator (T is the time between adjacent zero x-axis crossings).

I think you should be able to find T and thus your answer.

Note: in the case of underdamped motion like this problem, k/m is greater than (b/2m)^2. Also, realize that the "period" 2*T is not actually periodic - each period becomes smaller and smaller so only a given time period is useful. Hope that helps a little.