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 Jan 26th 2009, 12:31 AM #1 Junior Member   Join Date: Jan 2009 Posts: 9 error analysis the question is: when a body moves with velocity V1 for first half of the journey and V2 for the remaining journey, its average velocity is given by 2/average V = 1/V1 +1/V2. if V1 and V2 are (15 +- 0.5)m/s and (30 +- 0.1)m/s, find the average velocity within error limits.
Jan 26th 2009, 12:39 AM   #2
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 Originally Posted by the curious the question is: when a body moves with velocity V1 for first half of the journey and V2 for the remaining journey, its average velocity is given by 2/average V = 1/V1 +1/V2. if V1 and V2 are (15 +- 0.5)m/s and (30 +- 0.1)m/s, find the average velocity within error limits.
Which method of error propagation were you instructed to use?

 Jan 26th 2009, 11:41 AM #3 Senior Member   Join Date: Dec 2008 Location: Las Cruces NM Posts: 256 Perhaps your study materials already worked a problem with another quantity that was defined like the average velocity and you can imitate it (resistances in ohms or something like that). I suppose you could find the extremes of v by trying to solve for the min and max of $\displaystyle \frac {2 v_1 v_2} {(v_1 + v_2) }$ subject to the conditions that $\displaystyle 15 - 0.5 \le v_1 \le 15 + 0.5$ and $\displaystyle 30 - 05 \le v_2 \le 30 + 0.5$ Which would be a two variable calculus problem. That way assumes the 0.5 is an absolute bound. If 0.5 is the standard deviation of a normal distribution then we have a different problem.
 Jan 26th 2009, 09:07 PM #4 Junior Member   Join Date: Jan 2009 Posts: 9 combination of errors using the chain rule.
 Jan 26th 2009, 09:40 PM #5 Senior Member   Join Date: Dec 2008 Location: Las Cruces NM Posts: 256 That's a good idea, t.c. $\displaystyle V(v_1,v2) = \frac {2 v_1 v_2}{ v_1 + v_2}$ Compute $\displaystyle dV = \frac { \partial V} {\partial v1 } dv_1 + \frac {\partial V}{ \partial v_2} dv_2$ Let $\displaystyle dv_1 = 0.5$ $\displaystyle dv_2 = 0.5$ Evaluate the partial derivatives at $\displaystyle v_1 = 15$ $\displaystyle v_2 = 30$ Call the resulting value of $\displaystyle dV$ the "plus or minus" error in $\displaystyle V$.
 Jan 27th 2009, 09:33 PM #6 Junior Member   Join Date: Jan 2009 Posts: 9 Please explain the step marked as X in the following solution done by my teacher. Sol: V = average velocity; it is given that V = 2V1V2/ V1+V2 therefore V = 2 * 15 * 30 / 15 + 30 = 20 m/s. Now the error in V 2/V =1/V1 +1/V2 2dV/V^2 = dV1/V1^2 + dV2/V2^2 -------> X 2dV/400 = 0.5/225 + 0.1/900 dV = 0.46 so V = (20 +- 0.4)m/s.
 Jan 27th 2009, 10:27 PM #7 Member   Join Date: Nov 2008 Location: PAKISTAN Posts: 79 Newton Sorry, im busy wd students but let u know, latter if u dont find a sol. Bye
 Jan 27th 2009, 11:50 PM #8 Senior Member   Join Date: Dec 2008 Location: Las Cruces NM Posts: 256 I think the step is "implicit differentiation". We pretend V, V1 and V2 are all functions of some variable, which I will call x. In the equation $\displaystyle \frac 2 V = \frac 1 V_1 + \frac 1 V_2$ Take the derivative of both sides with repsect to x. The derivative of left side with respect to x is $\displaystyle 2(-\frac 1 V_2) dV$ where $\displaystyle dV$ is $\displaystyle \frac {dV}{dx}$. The derivative of the right hand side with respect to x is $\displaystyle (- \frac 1 V_1^2)dV1 + (- \frac 1 V_2^2)dV2$ where $\displaystyle dV1 = \frac {dV_1}{dx}$ and $\displaystyle dV2 = \frac {dV_2}{dx}$ All the derivatives involve he pattern $\displaystyle f(x) = \frac 1 {x^n} = x^{-n}$ By the chain rule, the derivative of $\displaystyle f(V(x)) = f'(V(x)) V'(x)$ $\displaystyle =( -n V(x)^{-n-1} )V'(x) = -n \frac 1 {V(x)^{n-1}} V'(x)$ $\displaystyle = -n \frac 1 {V(x)^{n-1}} dV$ So $\displaystyle 2(-\frac 1 V_2) dV = (- \frac 1 V_1^2)dV1 + (- \frac 1 V_2^2)dV2$ Multiply both sides by (-1) and you get the teacher's equation. The we solve that equation for $\displaystyle dV$

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