How Does Car Suspension Impact Vehicle Dynamics Over Humps?

[konichiwa3x]

In practice, the suspension system of a car consists of a spring under compression
combined with a shock absorber which damps the vertical oscillations of the car. Draw sketch graphs, one in each case, to illustrate how the vertical height of the car above the road will vary with time after the car has just passed over a hump if the shock absorber is:
(i) not functioning.
(ii) operating normally.

When the driver of mass 80 kg, steps into the car of mass 920 kg, the vertical height of the car above the road decreases by 2.0 cm. If the car is driven over a series of equally spaced humps, the amplitude of the vibration becomes much larger at one particular speed.
(i) Explain why this occurs. ( I understand that this happens due to resonance)
(ii) Calculate the separation of the humps if it occurs at a speed of 15 ms-1.

Please tell me the answers to these questions.

 

[tiny-tim]

Welcome to PF!

Hi konichiwa3x! Welcome to PF!

Show us what you've tried, and how far you've got … then we'll know how to help you!

 

[konichiwa3x]

hi, thanks for replying.
Well for the first part, I am not at all sure. This type of question is very new to me. Can you provide me a link where I can read up on it?

For the second part, (i) I understand that the amplitude of vibration becomes very large when the frequency of oscillation matches with the natural frequency (resonance).
(ii) I have calculated the spring constant and it comes out to be $$3.92$$x $$10^4$$. How do I proceed?

 

[tiny-tim]

I have calculated the spring constant and it comes out to be $$3.92$$x $$10^4$$. How do I proceed?

In other words: how do you find the frequency?

Well, you know that the acceleration of the end of a spring is minus the spring constant times the length of the spring.

So it's a harmonic (quadratic) equation, d^2x/dt^2 = -µx.

And the solution to that is … ? :smile

 

[konichiwa3x]

$$15 = (distance \quad between \quad humps)/T$$

where T is the time period of oscillation.

$$T = 2\pi/\omega= 2\pi \sqrt{\frac{m}{k}}$$

where 'm' is the mass of the man? (or is it the mass of man+car) Please explain.

And what about the first part of the question regarding the graphs?
Thanks for your time.

 

[konichiwa3x]

Please reply.

 

[tiny-tim]

$$15 = (distance \quad between \quad humps)/T$$

where T is the time period of oscillation.

$$T = 2\pi/\omega= 2\pi \sqrt{\frac{m}{k}}$$

where 'm' is the mass of the man? (or is it the mass of man+car) Please explain.

Hi konichiwa3x!

Sorry to take so long.

ok … you know the actual formula, so you don't need the quadratic equation … that saves time!

If your calculation of the spring constant is correct (and you haven't shown us, so I can't check it), then you must use the mass of man+car, because that's what the suspension is having to put up with!

And what about the first part of the question regarding the graphs?

"sketch graph" means that you don't need to put in any values … the examiner only wants to see the rough shape of the graph (eg, is it a line, a parabola, a sine curve, a V-shape, …).

So just say in words what the shape is … and then draw it!

To start you off … in words, what happens to the height of the car if there is no suspension (ie suspension not functioning)?