Thread: Suspension system/damping View Single Post
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 Quote by 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.
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)?