Jay K.

asked • 12/31/17

How to workout zeta in simple harmonic motions by only give time of oscillations.

A mass M of 5 kg is suspended as shown in Figure Q1 with a damper C and a spring K of stiffness 4000 N/m. When allowed to vibrate freely, four oscillations of the system take 0.90 s. Hi sorry to bother you, I'm just really confused on how to work out zeta with only given the time of the oscillations. I want to know how which formula to use by using the oscillations to get zeta. Thanks in advance

Arturo O.

What is zeta?  Do you mean the damping parameter?
Report

12/31/17

Arturo O.

Also, keep in mind when you post a question that we do not have the figures in your book.
Report

12/31/17

Jay K.

Yes damping parameter, damping factor same.
Report

12/31/17

1 Expert Answer

By:

Arturo O. answered • 12/31/17

Tutor
5.0 (66)

Experienced Physics Teacher for Physics Tutoring

See tutors like this
See tutors like this
In simple harmonic motion,
 
Md2x/dt2 + Cdx/dt + Kx = 0
 
This can also be written as
 
d2x/dt2 + (C/M)dx/dt + (K/M)x = 0
 
or
 
d2x/dt2 + 2ζω0dx/dt + ω02x = 0
 
Then
 
2ζω0 = C/M ⇒
 
ζ = C/(2Mω0)
 
You were given K, C, and M, and you know that 
 
ω0 = √(K/M).
 
Plug these into 
 
ζ = C/(2Mω0).
 
Upvote • 0 Downvote
More
Report

Jay K.

K and M were given but C hasnt been given it will be given after this question. i need to use the oscillations which has 4 over 0.90sec but I don't know how they get damping factor with that.
Report

12/31/17

Arturo O.

The oscillation information gives you a period.
 
T = (0.90 s)/4 = 0.225 s
 
You can get ω0 from
 
ω0 = 2π/T = 2π/(0.225 s) ≅ 27.93 rad/s
 
But
 
ω0 = √(K/M) = √(4000/5) rad/s ≅ 28.28 rad/s,
 
which is slightly different.  It looks like some of the numbers given in the problem are incompatible.  Also, without some indication of the damping strength (C), I see no way to find ζ.  Is anything else stated in the original problem?  And as I said before, we do not have figure Q1.
 
 
Report

12/31/17

Arturo O.

Another possibility that occurs to me is that the problem meant to say that the DAMPED system completes 4 cycles in 0.90 s (I originally interpreted the 0.90 s as referring to the undamped system), so the angular frequency of the damped system is
 
ω1 = 2π / [(0.90 s)/4] ≅ 27.925 rad/s
 
But it is also true that
 
ω12 = ω02 - (ζω0)2 = ω02(1 - ζ2)
 
So with a value for ω1, and knowing ωo = √(K/M), you can solve for ζ from the equation above.  Try it.  (And again, without figure Q1 to look at, it is hard to interpret the information in the wording of the problem.)
Report

01/01/18

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.