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Combining Functions problem

  • Thread starter KatelynO
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A point on the outside of the axle of a truck has a circular motion that can be modelled by a sine curve. If you measure the distance from the axle's centre to the bottom of the truck, that distance remains constant as long as the truck is on a level piece of road. If the truck goes over a bump, the springs absorb the shock but the truck bounces for a while. A truck is moving so that the axle, which is 6 cm in diameter, rotates at 1 rotation per second. As the truck hits a bump, the spring depresses by 20 cm and continues to depress by 80% of the previous bump as it bounces every half-second.
If the middle of the axle is 30 cm from the bottom of the truck, construct a graph illustrating the distance from a point on the outside of the axle to the bottom of the truck.


What is the equation of the graph?


To find the equation for this question I have to find two equations that meet all of the specifications of this question.
I found the period to be 1 second as the axle rotates once per second, and therefore can be represented by sinx. Then the truck hits a bump which causes the axle to raise 20cm (of a possible 30cm) towards the bottom of the truck, (I'm not sure about this part because it does not specify whether or not the spring then goes the 20cm opposite the bottom of the truck, towards the ground) with a continuous depression of 80% of the previous bump every half second.
f(x) = 20(0.8)^x(sinx)
This is the first time I've dealt with Combining Functions so I'm not very familiar with it, any help would be appreciated. Thank you
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi KatelynO! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)

(and why are you using x for time? what's wrong with t? :confused:)

The frequency of the bouncing is constant, so just find the amplitude, and multiply them (and finally add the axle rotation). :wink:
 
  • #3
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I've been working on this question a bit and I decided that
f(x)=30 - 20((0.80)^x)cos(2πx)+3sin(2πx) is a better representative of this question however now I am wondering how to graph cos(2pix) + 3sin(2pix) anyone have some tips for me?? It's the 2pi that's throwing me off. Thank you
 
  • #4
tiny-tim
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Hi KatelynO! :smile:

(have a pi: π :wink:)

Shouldn't the two periods be different?

Just draw the bigger wave in pencil, then "modulate" it with the smaller wave. :wink:
 
  • #5
SammyS
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... As the truck hits a bump, the spring depresses by 20 cm and continues to depress by 80% of the previous bump as it bounces every half-second. ...

... Then the truck hits a bump which causes the axle to raise 20cm (of a possible 30cm) towards the bottom of the truck, (I'm not sure about this part because it does not specify whether or not the spring then goes the 20cm opposite the bottom of the truck, towards the ground) with a continuous depression of 80% of the previous bump every half second. ...

Thank you
Hi Katelyn

(The discussion below refers to the distance between the bottom of the truck and the center of its axle.)

The description of the spring action suggests a spring action in which the distance between the bottom of the truck and the axle (center) oscillates at a constant rate, but with an amplitude which decays exponentially.

Model the exponential decay as A(t)=A(0)·e(-k·t), where t is the time, t=0 is the time at which the truck hit the bump. The initial oscillation starts out at 20 cm, when t=0, After on oscillation, 1/2 second, the oscillation is 80% of 20cm = (0.80)·(20cm)=16cm. Therefore, A(1/2)=0.16cm and A(0)=0.20cm.

Put these into equation: A(t)=A(0)·e(-k·t) and solve for k.

0.16(cm) = 0.20(cm)·e(-k·(0.5(s))

Divide both sides by 20(cm), take the natural log (ln) of both sides & solve for k.

How can you combine this with a sinusoidal oscillation?

Suppose the oscillation did not decrease with time. Then, starting with when the truck hit the bump, the bottom of the truck would be 10cm from the axle. The period of this oscillation is (1/2)(s) so, 1/4 (s) after hitting the bump, the bottom would be 50 cm (30+20) from the axle. etc.

Dbt-ca = 30 - (20·cos(2πt/.5)), where Dbt-ca is the distance from the bottom of the truck, to the center of the axle, and t is time in seconds.
 
  • #6
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Hi Katelyn

(The discussion below refers to the distance between the bottom of the truck and the center of its axle.)

The description of the spring action suggests a spring action in which the distance between the bottom of the truck and the axle (center) oscillates at a constant rate, but with an amplitude which decays exponentially.

Model the exponential decay as A(t)=A(0)·e(-k·t), where t is the time, t=0 is the time at which the truck hit the bump. The initial oscillation starts out at 20 cm, when t=0, After on oscillation, 1/2 second, the oscillation is 80% of 20cm = (0.80)·(20cm)=16cm. Therefore, A(1/2)=0.16cm and A(0)=0.20cm.

Put these into equation: A(t)=A(0)·e(-k·t) and solve for k.

0.16(cm) = 0.20(cm)·e(-k·(0.5(s))

Divide both sides by 20(cm), take the natural log (ln) of both sides & solve for k.

How can you combine this with a sinusoidal oscillation?

Suppose the oscillation did not decrease with time. Then, starting with when the truck hit the bump, the bottom of the truck would be 10cm from the axle. The period of this oscillation is (1/2)(s) so, 1/4 (s) after hitting the bump, the bottom would be 50 cm (30+20) from the axle. etc.

Dbt-ca = 30 - (20·cos(2πt/.5)), where Dbt-ca is the distance from the bottom of the truck, to the center of the axle, and t is time in seconds.
Hi Sammy and thank you for the help, unfortunately, the way you solved it is a bit over my head as I've never used natural logs or anything at that level. Is there any other way you could solve it maybe? Thank you again!
 

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