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# Simple Harmonic Motion

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1) A 1500 kg automobile sinks 6cm lower than its normal level when 4 passengers (each of mass 85kg) get in. Determine the frequency of oscillation of the atuomobile with the passengers on board, when it hits a bump on the road.

For this problem, I have to first find the spring constant of the spring and then use the formula f=1/(2pi) * square root of (k/m) to calculate the frequency.

But is the spring originally uncompressed when no passengers are in? (even though the car has a mass of 1500kg) Or is it already compressed at the beginning? If it's originally compressed, I wouldn't know the displacement from the uncompressed position and therefore I can't calculate the spring constant k.

If the spring is uncompressed when no passengers are in, I can say that the new equilibrium position with passengers in is x=-0.06m. (with x=0 representating the unstrecthed position)

Fnet=0

-mg+(-kx)=0

-340(9.8)-k(-0.06)=0

k=55533.33N/m

But then whe I substitute into f=1/(2pi) * square root of (k/m), which mass should I use, the mass of the car or the passengers?

Can someone explain? Thank you! :confused:

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eeeek! sitting on a goldmine with this post!

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But is the spring originally uncompressed when no passengers are in?
Of course not! But you can suppose linearity (Hooke's law).
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Allow me to do so. I have got 98.97% in an exam held on this just last year.

But is the spring originally uncompressed when no passengers are in? (even though the car has a mass of 1500kg) Or is it already compressed at the beginning? If it's originally compressed, I wouldn't know the displacement from the uncompressed position and therefore I can't calculate the spring constant k.

Though it is not actually uncompressed, you can assume that it is.

(Unless you want ultra accurate answers, and I mean till nanometers)

But then whe I substitute into f=1/(2pi) * square root of (k/m), which mass should I use, the mass of the car or the passengers?

Use the mass of the car summed to mass of the passengers.

Rember that mass is the inertia factor. We don't care about gravity in this problem.

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Though it is not actually uncompressed, you can assume that it is.

(Unless you want ultra accurate answers, and I mean till nanometers)

Like Qfwfq was trying to say it doesnt matter if it is or is not compressed, there is a linear relationship between the two, so it doesnt matter(in this case) if you are compressing from 0 to x or from y to y+x. Its all the same thing.

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Like Qfwfq was trying to say it doesnt matter if it is or is not compressed, there is a linear relationship between the two, so it doesnt matter(in this case) if you are compressing from 0 to x or from y to y+x. Its all the same thing.

Oh, so that's what linearity meant...:lol: Sorry, Qfwfq.

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dont worry about it :lol: I knew what he meant but I have never heard that word used before now!

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linear: y = mx + q

linear and homogeneous: y = mx

non linear: y =

a x ^2 + b x + c

x^3

x^(-1)

x^(1/2)

etc..........

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