How Do I Calculate the Natural Frequency and Time Elapsed for a Gas Compressor?

[Renni202]

I hope I am posting in the right place. I am a 1st year University student. Although I am undertaking a BSc in Computer and Electronic System Engineering I have a Mechanical Principles module which I am finding a little bit of a struggle.

Mechanical Principles - Dynamics:

a) A diaphragm type gas compressor of mass 235 kg is supported by 4 springs which are arranged in parallel. The springs each have stiffnesses of 40 kN/m, 40 kN/m, 45 kN/m and 45 kN/m respectively. Determine:

i) the total stiffness of the spring system in N/m;

ii) the natural frequency of free vibration of the machine in Hz;

iii) the time elapsed after 5 cycles of free vibration, and show this on an amplitude-time graph.Answer
i) K_T = 2K₁+2K₂
K_T=170x10³ N/m
i.e K_T=170 kN/m
(It askes for the answer in N/m so should I still take it to kN/m or stop at N/m?)

ii) f_n=1/2πK/m=ω_n/2π
=1/2π√170x10³/235
=4.2 Hz

iii) T=1/f_n=1/4.2=0.24s for 1 cycle 1.2 for 5 cycles

I am looking for confirmation that I am on the right track with my answers and any guidance offered on making an amplitude-time graph. It really is much appreciated.

 

[Simon Bridge]

Welcome to PF;

Answer
i) K_T = 2K₁+2K₂
K_T=170x10³ N/m
i.e K_T=170 kN/m
(It askes for the answer in N/m so should I still take it to kN/m or stop at N/m?)

It won't make any difference so long as you have the answer they wanted.
The question expected you to add up the kN values and they were making sure you knew what that meant.

ii) f_n=1/2πK/m=ω_n/2π
=1/2π√170x10³/235
=4.2 Hz

It looks like you applied an equation without understanding it.
That is not best practice. However, this will get you the marks.
You should include a note in words to make your math clear - like commenting code.

iii) T=1/f_n=1/4.2=0.24s for 1 cycle 1.2 for 5 cycles

I am looking for confirmation that I am on the right track with my answers and any guidance offered on making an amplitude-time graph. It really is much appreciated.

If you understood the relations you probably wouldn't need the confirmation, or you'd be able to do your own.

What sort of motion is this going to be?
From your knowledge of this sort of motion, how does the amplitude of the oscillations vary with time?

Note: do they want amplitude vs time, or the displacement of the machine (from equilibrium) vs time?

 

[Renni202]

Thank you Simon,

So thumbs up for part (i) YAY!

ii) ωn/2π = 26.8/2π=4.2 Hz I think I did miss out a step because my first answer was 42. I checked with a rad/s to Hz converter which gave me 4.2653524797 (I should actually round up to 4.3 I guess) To match the answer I used 26.8/6.283185 and assumed there was something going on with my calculator.

iii) The graph is going to be an up and down wave like form (I think). Each cycle should have a wave. Although I was unsure how high or low it should go. It confused me slightly because it was written... Amplitude - time graph not amplitude vs time or displacement. Am I being silly and not reading it correctly?

I have just realized the time here in the UK so I might need to pick this up tomorrow. Thank you again. Your reply has already given me a boost in confidence.

 

[Simon Bridge]

I don't look at arithmetic so if you bunged in the wrong numbers...

$$f=\frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{T}$$ 42 is, of course, the Ultimate Answer ... the 4.2 coming out suggests you misplaced a factor of 100.

Also see:
http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html

The natural frequency is what it would oscillate at without any damping or driving.
i.e. Simple Harmonic Motion.

In Simple Harmonic Motion, the amplitude is a constant over time.
If the vertical displacement from equilibrium is at time t is y(t), then

$y(t)=A\cos(\omega t+\delta)$

the amplitude is $A$.

See how an A vs t graph would just be flat? So the distinction is important.

There are two ways this can be not what they mean:
1. they actually mean you to sketch a y vs t graph (likely).
2. the context is not simple harmonic motion -
...i.e. there may be damping so the amplitude decays exponentially with time.

You should be able to sketch a sine wave.
In one period, you get one peak and one trough.

 

[Renni202]

I had the image of a sine wave in my head but was not sure exactly what I was looking for. Your a star thank you. I have a couple of other questions I plan on working through tomorrow. Will it be ok to add them to this post? (Same subject. Same coursework sheet.)

 

[Simon Bridge]

No worries.
It is best to make a separate thread for each new question.