Vibrations - Modeling system, equation of motion

[haruspex]

Haha I had the minus in my first answer and you said it was wrong ? That's why I'm confused

There are two terms in the factor for θ. Originally you had both negative, which will not give SHM. The overall value must be positive (ω²). Now you have both positive, yet you stated, correctly, that one increases while the other decreases.

And for deducing ω from the graph, I see that θ = 0.02 at ωt=0, but how do i get ω from that?

What's the period?

 

[thepikminman]

There are two terms in the factor for θ. Originally you had both negative, which will not give SHM. The overall value must be positive (ω²). Now you have both positive, yet you stated, correctly, that one increases while the other decreases.

What's the period?

Okay, I get it now, thank you for your patience! So it's: θ''I_n + 0θ' + (m₂gl₂ - kl₃) θ= 0

As for the period...I know it's = 1/f = 2π/ω, but I don't know what f is. The period seems to be 6 from the graph but that's just by looking, so is it 1/6?

 

[haruspex]

how can I get the sqaure root of 4*[m1l12 + m2l22 + m3l32]*(m2gl2 + kl3)?

I just noticed that all the constants are given in Table 2 at the top of page 4, so you can calculate this.

θ''In + 0θ' + (m2gl2 - kl3) θ= 0

You wrote, correctly, that as θ increases the effect of the spring (kl₃θ) is to reduce $\ddot\theta$. So what sign should the kl₃θ term have? You might find it easier to answer if you rearrange the equation as $\ddot θ$I_n = (+ or -?)m₂gl₂θ + (+ or -?)kl₃ θ.

 

[thepikminman]

I just noticed that all the constants are given in Table 2 at the top of page 4, so you can calculate this.

You wrote, correctly, that as θ increases the effect of the spring (kl₃θ) is to reduce $\ddot\theta$. So what sign should the kl₃θ term have? You might find it easier to answer if you rearrange the equation as $\ddot θ$I_n = (+ or -?)m₂gl₂θ + (+ or -?.

Ah yes, the table, that makes sense!

So it would be θ''I_n = (+m₂gl₂ - kl₃ )θ...right?

 

[haruspex]

Ah yes, the table, that makes sense!

So it would be θ''I_n = (+m₂gl₂ - kl₃ )θ...right?

Yes.

 

[thepikminman]

Yes.

Okay, so that gives me the correct DE, now is finding the roots of the characteristic equation the next step? (with b=0)

 

[haruspex]

Okay, so that gives me the correct DE, now is finding the roots of the characteristic equation the next step? (with b=0)

As I posted, it is easier to work from the generic equation $\theta=A\cos(\omega t+\phi)$. You can calculate ω from the coefficient in the differential equation $(\ddot \theta+\omega^2\theta=0)$, and read off the amplitude and φ from the graph.

 

[thepikminman]

As I posted, it is easier to work from the generic equation $\theta=A\cos(\omega t+\phi)$. You can calculate ω from the coefficient in the differential equation $(\ddot \theta+\omega^2\theta=0)$, and read off the amplitude and φ from the graph.

Is the coefficient of the θ variable always = ω²?

If so, can't I just get ω² = √27.513 (the coefficient in front of θ) = 5.245?

Is that the natural frequency value?

How do I go about reading the amplitude and φ from the graph anyway? A seems to be about 3.9 (?) but how do I read φ ? The sin wave seems to be shifted 5.5 rad to the right so is θ = 3.9sin(ω(0)-5.5)? (⇒ θ = 2.75 rad?

 

[haruspex]

Is the coefficient of the θ variable always = ω2?

Yes. Plug θ=A cos(ωt+φ) into $\ddot \theta+\omega^2\theta$ and see what happens.

27.513

If that is = (+m₂gl₂ - kl₃ )/I_n, yes.

 

[haruspex]

How do I go about reading the amplitude and φ from the graph

You could measure with a ruler, but by eye I would put the amplitude about 0.037 or 0.038.
For the phase, the first peak is at about ωt=0.8 or 0.9 radians. What value of φ would result in that?

 

[thepikminman]

Yes. Plug θ=A cos(ωt+φ) into $\ddot \theta+\omega^2\theta$ and see what happens.

If that is = (+m₂gl₂ - kl₃ )/I_n, yes.

Ah they cancel, that's cool! But does that only work for this particular example? With the θ' coefficient being = 0?

Basically, I'm asking if that's a way to calculate ω in every question, not just this one? Because most past exam questions ask for ω_n, so if I find the equation of motion, then will ω_n always be the coefficient of the "x" term (θ in this problem)?

 

[thepikminman]

You could measure with a ruler, but by eye I would put the amplitude about 0.037 or 0.038.
For the phase, the first peak is at about ωt=0.8 or 0.9 radians. What value of φ would result in that?

Would it be sin(θ-0.9), so ψ = 0.9?

On a different topic, I'm able to do most of the questions now, but I have just 2 last queries for you;

1. In Question 6(a) I derived the formula fine, but what are the units for ω_n in terms of E,I,L and m? I did it as ω_n = √(Gpa*mm⁴)/(mm³*kg) = √(Gpa*mm)/kg...does that results in the correct units for the natural frequency? If not, what does?

2.In Question 6(c) how does the factor of safety come into play here? The "hint" with λ in it actually made me think I have no idea how to answer this correctly because none of my lecture notes have that info on modes or λ(c). Any ideas on how to approach this?

Thanks so much to you and @BvU for all your help so far, you really helped me take the edge off this difficult subject.

 

[haruspex]

if that's a way to calculate ω in every question,

Yes, but don't forget the square root. If $\ddot x + kx=0$ then the angular frequency is $\sqrt k$.
(If k is negative it is not SHM, since the system will accelerate away from the neutral position, like a pencil balanced on its point).

Would it be sin(θ-0.9), so ψ = 0.9?

Careful with signs. Substitute that in the equation with ωt=0.9 to check.

√(Gpa*mm)/kg...does that results in the correct units for the natural frequency?

It certainly produces the right dimension (1/time). Do you know dimensional analysis? But you need to check whether the units are s^-1.

how does the factor of safety come into play here?

This gets beyond my competence. Knowing the natural frequency etc. and the applied frequency and power, there must be some formula that tells you the maximum stress. Look up forced oscillations.

The "hint" with λ in it

All this is saying is that the mode is the simplest possible: at any instant, the displacement at offset x is c(x)cos(ωt), where c(x) gives the shape of the beam under a static deflection.

 

[thepikminman]

Careful with signs. Substitute that in the equation with ωt=0.9 to check.

Thank you!

So it the angular displacement, θ = Acos(ωt + Φ) = 0.037cos(0 + 0.9)?

 

[haruspex]

Thank you!

So it the angular displacement, θ = Acos(ωt + Φ) = 0.037cos(0 + 0.9)?

No, you misunderstand.
According to the graph, you need a peak at ωt=0.9. What value of Φ maximises Acos(0.9 + Φ)?

 

[thepikminman]

No, you misunderstand.
According to the graph, you need a peak at ωt=0.9. What value of Φ maximises Acos(0.9 + Φ)?

Oh, is it for Φ=-0.9, becuase cos(0) is the maximum value you can get?

 

[haruspex]

Oh, is it for Φ=-0.9, becuase cos(0) is the maximum value you can get?

Yes.