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Physics please help Will give Best Answer Picture in details! Your car rides on springs, so it will have a natural frequency of oscillation?
Your car rides on springs, so it will have a natural frequency of oscillation. (Figure 1) shows data for the amplitude of motion of a car driven at different frequencies. The car is driven at 30 mph over a washboard road with bumps spaced 13 feet apart; the resulting ride is quite bouncy.
Determine the frequency of the oscillation, caused by the bumps. 1 mile is 5280 feet
PICTURE https://session.masteringphysics.com/problemAsset/...
Determine the frequency of the oscillation, caused by the bumps. 1 mile is 5280 feet
2 Answers
- Andrew SmithLv 74 years agoFavorite Answer
As the bumps are 13 feet apart the wavelength = 13 feet
V = f * lambda
f = V /lambda
But V must match the units of wavelength. So convert V to feet per second
V = 30 * 5280/3600 feet per sec
-> f = (30 * 5280/3600)/ 13 = 3.4 Hz
This is the frequency of the bumps on the road.
From the graph we can see that the frequency is around 2 Hz so one concludes that the frequency of the car is actually f/2 and each bump coincides with each SECOND bounce of the car.
ie the wavelength is actually 13/2
Now the only trouble is that if this was true then there should be a second peak at f = 3.4 Hz on the graph.
As this does not occur the graph appears to be inconsistent with the data of the question.
