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1. When a student pushes down with a force of 200N on the wing of a car above one of its front wheels, the wing is displaced downwards by 12mm.
a) Calculate the stiffness (i.e. spring constant) k of the suspension spring at that wheel.
b) Assuming each wheel is fitted with an identical spring and the mass of the car is 600kg (with driver), calculate its natural frequency of oscillation.
c) Estimate the speed at which the car would resonate when travelling over regularly spaced bumps 15m apart.

parts b and c i'm stuck on.
Thanks.

b) 1.68Hz
c) 25.1 m/s
2. b) Assuming each wheel is fitted with an identical spring and the mass of the car is 600kg (with driver), calculate its natural frequency of oscillation.

natural frequency = (1/2pi).(g/L)^1/2

L is going to be 12mm = 12 x 10^-3m

So natural frequency of oscillation = 1/2pi . (9.81/12x10^-3)^1/2
= 28.7/2pi... approx (not got a calculator to hand)

c) Estimate the speed at which the car would resonate when travelling over regularly spaced bumps 15m apart.
v = f.> where > is wavelength
v = 15 x 28.7/2pi

WTF. Sorry, no idea for c, where on Earth did you find such a question... That is really awkward.
3. 1.68Hz is the resonant frequency

the frequency of the bumps which are 15 m apart, for a car travelling at v is given by v/15 (this is clear by considering units)

so 1.68 = v/15
v = 25.2 m s^-1
4. (Original post by mik1w)
1.68Hz is the resonant frequency

the frequency of the bumps which are 15 m apart, for a car travelling at v is given by v/15 (this is clear by considering units)

so 1.68 = v/15
v = 25.2 m s^-1
could you calculate the resonant frequency for me?
5. T = 2*pi*sqrt(m/k) for the oscillations of a mass spring system

you know m = 600 kg
and k = 1670 N/m
from you own calculations

And also that f = 2*pi/T, so from a simple substitution, f = sqrt(k/m)

f = sqrt(1670/600)
f = 1.67 Hz
6. (Original post by mik1w)
T = 2*pi*sqrt(m/k) for the oscillations of a mass spring system

you know m = 600 kg
and k = 1670 N/m
from you own calculations

And also that f = 2*pi/T, so from a simple substitution, f = sqrt(k/m)

f = sqrt(1670/600)
f = 1.67 Hz
surely k = 16700N/m?
7. When a student pushes down with a force of 200N on the wing of a car above one of its front wheels, the wing is displaced downwards by 12mm.k = F/x = 200 / 0.012 = 16700, yes that's rtue

so the *******s have got it wrong again?

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Updated: November 24, 2005
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