Solving for the Damping Constant

  • #1
lightlightsup
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Homework Statement
The suspension system of a mass ##M## automobile "sags” a distance ##x## when the chassis is placed on it. Also, the oscillation amplitude decreases by a factor ##N## each cycle. Estimate the values of (a) the spring constant ##k## and (b) the damping constant ##b## for the spring and shock absorber system of one wheel, assuming each wheel supports an equal fraction of the weight. Make the simplifying assumption that the change in the period due to damping can be ignored. Express your answer in terms of the variables given and ##g##.
Relevant Equations
See attempt.
##x(t)=x_m e^{\frac{-bt}{2m}}cos(ωt+φ)##
Damping Factor: ##e^{\frac{-bt}{2m}}## (##b## is the damping constant)
Estimated ##ω##: ##\sqrt{\frac{k}{m}}##
More Accurate ##ω'##: ##\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}##
Also: ##T = \frac{2π}{ω}## and ##F = kx##.

So, my answers:
##k = \frac{Mg}{4x}##
Estimate: ## b = -\frac{M\ln(N)}{4π}\sqrt{\frac{g}{x}}##
More Accurate: ## b = \sqrt{\frac{gM^2\ln^2(N)}{x[16π^2+4\ln^2(N)]}}##

Apparently, my answer for ##k## is correct but my answer for ##b## is wrong.

If you need numbers to test these equations:
••60 The suspension system of a 2000 kg automobile “sags” 10 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 50% each cycle. Estimate the values of (a) the spring constant ##k## and (b) the damping constant ##b## for the spring and shock absorber system of one wheel, assuming each wheel supports 500 kg.
The answers are: (a) 49,000 N/m, (b) ~1100kg/s (1092 with the estimated ##b## and 1086 with the more accurate ##b## formula).

This is from Halliday's Fundamentals of Physics, 11th Edition, Chapter 15 (Oscillations).
 
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  • #2
lightlightsup said:
So, my answers:
##k = \frac{Mg}{4x}##
Estimate: ## b = -\frac{M\ln(N)}{4π}\sqrt{\frac{g}{x}}##
More Accurate: ## b = \sqrt{\frac{gM^2\ln^2(N)}{x[16π^2+4\ln^2(N)]}}##

Apparently, my answer for ##k## is correct but my answer for ##b## is wrong?
Why do you say your answer for ##b## is wrong. Your formula for the "estimate" and for the "more accurate" look correct.
 
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  • #3
TSny said:
Why do you say your answer for ##b## is wrong. Your formula for the "estimate" and for the "more accurate" look correct.

Thank You.
I thought so.
I think Wiley (the publisher) is wrong.
Their online homework system does not tell me why I'm wrong.

My Answer in JS. Change the variable values on line 1.:
M = 2000; x = 0.1; N = 0.5; g = 9.8;
k = ((0.25*M)*g)/x;
b_simple = (((-1*(0.25*M))*Math.log(N))/(Math.PI))*(Math.sqrt(g/x));
b_complex = Math.sqrt((g*(M**2)*((Math.log(N))**2))/(x*((16*((Math.PI)**2))+(4*((Math.log(N))**2)))));
console.log("k =",k);
console.log("b (simple) =",b_simple);
console.log("b (complex) =",b_complex);
 
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  • #4
OK.

lightlightsup said:
If you need numbers to test these equations:
••60 The suspension system of a 2000 kg automobile “sags” 10 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 50% each cycle. Estimate the values of (a) the spring constant ##k## and (b) the damping constant ##b## for the spring and shock absorber system of one wheel, assuming each wheel supports 500 kg.
The answers are: (a) 49,000 N/m, (b) ~1100kg/s (1092 with the estimated ##b## and 1086 with the more accurate ##b## formula).

This is from Halliday's Fundamentals of Physics, 11th Edition, Chapter 15 (Oscillations).
Is the above quote taken from the textbook? I'm curious as to why the answer for ##b## is given as a negative number (-1100 kg/s). From your expression for ##x(t)##, ##b## should be a positive number.
 
  • #5
TSny said:
OK.
Is the above quote taken from the textbook? I'm curious as to why the answer for ##b## is given as a negative number (-1100 kg/s). From your expression for ##x(t)##, ##b## should be a positive number.

That's a ~, not a -.
 
  • #6
lightlightsup said:
That's a ~, not a -.
What is the meaning of ~ ? Is it supposed to be an "approximate" symbol?
 
  • #8
lightlightsup said:
OK. For some reason, the symbol is displaying as a hyphen on my screen. Thanks.

(Interesting: In Chrome it displays as a hyphen, but in Explorer it displays correctly as ##\sim##)
 
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  • #9
This is the solution, as per WileyPlus. Does this make sense?
I fail to see a scenario where b would not evaluate to a negative number.
There should be a negative sign in the answer to turn the ##\ln{N}## into a positive?
Wiley Solution.png
 
  • #10
The original problem statement says ##N## is the factor by which the amplitude decreases per cycle. This could be interpreted as saying that ##N## is what you multiply the initial amplitude by in order to get the next amplitude. In this case, ##N## is less than 1. This leads to your formula for ##b## which has the overall negative sign outside that combines with the negative value of ##ln N## so that ##b## is positive.

WileyPlus takes ##N## to be a number that you divide the initial amplitude by. So, now ##N > 1##. The formula for ##b## will no longer have an overall negative sign. But ##b## will still come out positive.
 
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  • #11
TSny said:
The original problem statement says ##N## is the factor by which the amplitude decreases per cycle. This could be interpreted as saying that ##N## is what you multiply the initial amplitude by in order to get the next amplitude. In this case, ##N## is less than 1. This leads to your formula for ##b## which has the overall negative sign outside that combines with the negative value of ##ln N## so that ##b## is positive.

WileyPlus takes ##N## to be a number that you divide the initial amplitude by. So, now ##N > 1##. The formula for ##b## will no longer have an overall negative sign. But ##b## will still come out positive.

Wish they were clearer about this.
Guess I have to brush up on my math and interpretation game.
Thank You.
 
  • #12
lightlightsup said:
Wish they were clearer about this.
Guess I have to brush up on my math and interpretation game.
Thank You.
They should definitely have been clearer on the interpretation of N. Especially if they were accepting only one interpretation as the correct answer.
 
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