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SHM

Hi I'm having prolems getting my head around the maths of damped simple harmonic motion.

E.O.M: xdoubledot + gamma*xdot + omeganought^2 * x = 0 (correct?)

Is the natural frequency omega nought? does it equal 2*pi*f, and if not how do i calculate it without the spring consant k?

This is the question:

A 20 kg mass on a spring exhibits damped simple harmonic motion. The mass
is released from an initial displacement of 5 cm and is observed to oscillate with
a period of 5 s. Five seconds later its maximum displacement is observed to be
1.84 cm. Calculate:

(i) the damping constant of the system
(ii) the natural frequency of the system (i.e. the frequency that would be seen in the absence of damping)
(iii) the tension in the spring five seconds after the mass is released

I'm having trouble with all parts of this question

(edited 13 years ago)

Reply 1

Stonebridge

Original post by samueltaylor123

I think all you need is here
http://en.wikipedia.org/wiki/Damping#Under-damping_.280_.E2.89.A4_.CE.B6_.3C_1.29
The system is under-damped.
The damping ratio is found from the log of the ratio of the two amplitudes given. Two successive amplitudes.
The damped frequency of oscillation (given) is related to the undamped frequency by the equation given in the link.

w_d = w_0\sqrt{1-\zeta^2}

The undamped frequency of oscillation is also given by

w_0 = \sqrt{\frac{k}{m}}

from which you can find the spring constant, k.

(edited 13 years ago)

Reply 2

samueltaylor123

whats zeta in the square root?

Reply 3

Stonebridge

Original post by samueltaylor123

whats zeta in the square root?

It's the damping ratio, and is related to the damping coefficient c through an equation you will find at the top of the page I linked to.

\zeta = \frac {c}{2m w_0}

I assume that's what your question is asking for when it says damping constant. It's the viscous damping coefficient that determines the magnitude of the damping force.
Zeta is a constant that describes the resulting motion. In particular, the rate at which the amplitude decays.
In this question, zeta<1 and the motion is under-damped.

(edited 13 years ago)

Reply 4

samueltaylor123

ok thank you. we havent been taught the zeta function.

Reply 5

samueltaylor123

and what about the tension? I was thinking use

F=-kx where k can be found using omega and x is the displacement of 1.84cm

so this is the spring restoring force acting upwards, but would I have to include gravity acting down?

Reply 6

kaosu_souzousha

For the last one, I guess you just refer to our mate Newton and his wonderful second law of motion. Tension will be maximum at the bottom.
F= kx (tension acting upwards and weight acting downwards)
so kx = mg
I hope I am not missing anything important

Reply 7

Stonebridge

Original post by samueltaylor123

Yes tension is just k times extension but you need to be careful with the value of the extension.
When the amplitude is a, the spring is extended a + x where x is the extension in equilibrium. That is, when the mass is hanging motionless and the tension is mg.
So yes, you have to add mg to the tension you calculate from k times the amplitude.