Amplitude problem — Car shock absorbers going over a bump

[robax25]

Homework Statement

-(View image attached)[/B]
A driver drives a car.The road has periodic small bump of 5 cm height and a distance of 5 cm. Car's shock observer works fine,damping the deflection to half each oscillation.Their length is 20cm increased if car weight(1500kg) does not act.Determine the oscillation amplitude if the car is driving at 20km/h.

consider the problem by assuming that the distance of the four wheels equals multiple of the periodicity.

Homework Equations

X(amplitude)=Xoe^(-bt/2*m)
E=.5 kx² e^-(bt/m)
damping force Fd=-bv

The Attempt at a Solution

x=10cm[/B]

 

[haruspex]

3. The Attempt at a Solution
x=10cm

Please post your working.

 

[robax25]

my solution is wrong and I need to know right way how to get exact solution.

 

[haruspex]

my solution is wrong and I need to know right way how to get exact solution.

We cannot tell where you have gone wrong if you do not post your working.

 

[robax25]

can you give me any idea how i can proceed?

 

[robax25]

My first assumption was wrong. So I did so far. f=v/λ = 20km/h/0.2m =27.8 HZ SO ω=2*η*f = 174.7 rad/s

 

[robax25]

Actually I thought the amlitude is =5+5=10cm but it is wrong

 

[haruspex]

20km/h/0.2m =27.8 HZ

0.2m? You originally stated 0.05m (which did strike me as rather small).

 

[robax25]

Their length is 20cm increased if car weight(1500kg) does not act. Their length means oscillation(wave length) length

 

[haruspex]

Their length is 20cm increased if car weight(1500kg) does not act. Their length means oscillation(wave length) length

I think you are confusing the two directions. You divided the speed of the car by a distance, so that should be a horizontal distance, namely, the distance between bumps. That cannot change. What would be increased without the car's mass is the vertical deflection.

 

[robax25]

amplitude 20 cm is increased if car weight does not act

 

[robax25]

0.2m? You originally stated 0.05m (which did strike me as rather small).

it should be 20km/h/0.05m =111.11hz and omega = 698.13 rad /s

 

[robax25]

Now how can I find out amplitude?

 

[haruspex]

Now how can I find out amplitude?

There are three amplitudes and you are told all of them:

bump of 5 cm height

(what amplitude does that imply?)

damping the deflection to half

(what amplitude now?)

length is 20cm increased if car weight(1500kg) does not act

I'm not quite sure how to use this last fact.
First, is that increased by 20cm or to 20cm?
Secondly, I don't think you can get a meaningful equation with no mass involved, so my guess is that there is an unknown mass of the shock absorber. With that mass alone the amplitude is 20cm (or 20+?); with the added 1500kg it is much less.

The situation described is a forced-damped oscillation. What equations do you know for that?

 

[robax25]

for bump, the Amplitude is 5cm and for deflection, it is now 2.5 cm and last one is force damped oscillation.

 

[robax25]

forced damping amplitude A=F/[m²(ωo² - ω²)²+b²ωo²]^1/2
b is the damping constant

 

[haruspex]

Amplitude is 5cm

The bumps have height 5cm. Amplitude is the deflection each side from a central position.

 

[haruspex]

forced damping amplitude A=F/[m²(ωo² - ω²)²+b²ωo²]^1/2
b is the damping constant

Right.
Can you apply that equation to the two different circumstances described in the question, i.e. with and without the cars's mass?
(I think I was wrong in saying you need an unknown for the spring's mass. You can take that as zero.)

 

[robax25]

The bumps have height 5cm. Amplitude is the deflection each side from a central position.

that means amplitude is 5 cm for bump and for energy loss (50%) it reduces its Amplitude. am I right or wrong for bump amplitude?

 

[robax25]

Their length is 20cm increased if car weight(1500kg) does not act. I do not understood how to use 20cm to get amplitude. 20 cm is also amplitude if you drive the car near to natural frequency. That is why it gets suddenly higher from 5cm to 20cm. you are right I have to use two cases when mass is 0 and when mass is 1500. But how to get driving force?

 

[haruspex]

am I right or wrong for bump amplitude?

Wrong. 5cm is the height of the bumps, i.e. the vertical distance from the bottom of a bump to a top. These are the extremes of oscillation of the bumps. Amplitude of oscillation is not the distance between extreme displacements; it is the distance from a central position to each extreme.

I do not understood how to use 20cm to get amplitude.

I find the wording vague. What are 'they' in "their length", how is their length defined, and is it increased by 20cm or to 20cm?
Perhaps 'they' are the oscillations of the top of the shock absorber, length is amplitude and it is increased by 20cm, but there are other possibilities.

 

[robax25]

I asked professor how it works. Their length 20cm is increased means when weight(1500 kg) of the car acts, the spring compresses to 20 cm and when the weight(1500 kg) does not act, the spring returns to equilibrium position.

 

[robax25]

so we can Fo= mg , so we get damping force.

 

[haruspex]

I asked professor how it works. Their length 20cm is increased means when weight(1500 kg) of the car acts, the spring compresses to 20 cm and when the weight(1500 kg) does not act, the spring returns to equilibrium position.

Ah, ok. So that is just telling you the spring constant.

so we can Fo= mg , so we get damping force.

Not damping. As you posted at the start, damping force is related to velocity (in the direction of the force), not displacement.
Indeed, I'm not sure that this question involves damping at all. It is a forced oscillation (the bumps in the road) acting on a spring. What is the general equation for that?

 

[robax25]

I asked him He told me that it is forced oscillation problem. forced damping amplitude A=F/[m²(ωo² - ω²)²+b²ωo²]^1/2
b is the damping constant

 

[haruspex]

He told me that it is forced oscillation problem.

Yes, that's what I wrote. But is he saying it is also damped, or did you add that?

Edit: the isuue is, do we have enough information to solve it for damping as well as forcing. Let's suppose we do and we can drop the damping if it turns out we do not.

 

[robax25]

By reading the question, it is completely clear that there is damping. The statement is "A driver drives a car.The road has periodic small bump of 5 cm height and a distance of 5 cm. Car's shock observer works fine,damping the deflection to half each oscillation".

 

[haruspex]

By reading the question, it is completely clear that there is damping. The statement is "A driver drives a car.The road has periodic small bump of 5 cm height and a distance of 5 cm. Car's shock observer works fine,damping the deflection to half each oscillation".

The use of the word damping there should not be trusted. It means merely that the oscillation is reduced to half. That can happen by virtue of a mismatch between the frequency of the forcing and the natural frequency of the spring. A real suspension spring damps too, but I'm not yet sure that we have enough information to allow for that. As I posted, let's suppose we do for now.

 

[haruspex]

forced damping amplitude A=F/[m²(ωo² - ω²)²+b²ωo²]^1/2
b is the damping constant

Ok. So define all those variables.

 

[robax25]

A=amplitude, F=force acts on the spring...ω0=initial angular velociy, and ω'= damped angular velocity However, I do not have damping constant.If we consider it is undamped oscillation then we can get ω0=ω'

 

[robax25]

According to the question we can write also e^bt/2m=1/4

 

[haruspex]

F=force acts on the spring

No, it can't be that. To fit in dimensionally with the rest of your equation it has to be an acceleration, not a force. Also, the rest of that equation is time-independent, so it cannot be anything that varies over time.
I believe it should be the driving amplitude (resulting from the bumps in the road) multiplied by the spring constant.
See http://farside.ph.utexas.edu/teaching/315/Waves/node13.html for a suitable model for this problem.

ω'= damped angular velocity

Your equation has ω, not ω'. They are different.

If the height of each bump is 5cm, what amplitude of driving oscillation is that?
What is the value of the spring constant?
What is the undamped frequency (ω) of the spring?
All of those can be deduced from the given information.

 

[robax25]

the amplitude of driving oscillation is 2.5cm
spring constant k=73575N
ωo(natural frequency)=7 rad/s
m=1500kg
According to your link,
X0=ωo²Xo/√((ωo²-ω²)²+v²ω²) v= damping constant
v=ωom/(2π*m/ΔE/E)
v=7rad/s*1500kg/(2π*1500kg/0.5) As energy loss is 50%:

v=0.557 kg/s

ωo=ω as driven frequency X0=ωo²Xo/vωo
= 31.4cm

 

[haruspex]

Car's shock observer works fine,damping the deflection to half each oscillation

It has at last dawned on me what this is saying.
It means that if the car were standing still and you were to push it down by 10cm and let go then it would bounce up and down, with the next down being only 5cm below equilibrium. From this you can calculate the damping constant (but it is not trivial).

I think it is clear that the answer to the question should be significantly less than the height of the bumps, so no more than 1mm. (I calculate it to be a lot less.)

 

[robax25]

I did not consider speed of the car, it has to consider. I will ask him again.