Question

In this problem, you are going to calculate the maximum possible error and uncertainty for a spring that is used to measure force. The spring is considered linear, so that F = kx, where F is force in newtons, k is the spring constant in newtons per cm, and x is the displacement in cm. If x = 15.0 ± 0.15 cm , and k = 1500 ± 15.0 N/cm, calculate the maximum possible error and the reasonable uncertainty of the measured force in absolute (dimensional) and relative (%) terms. Select the most correct response (a, b, ... h)

Answer 1

Answer:

Maximum possible error is ±450

Uncertainty in absolute terms is ±318.1981

Relative terms is 1.4%

Explanation:

Spring constant, $k±w_{k}$ is 1500±15 N/cm and displacement, $x±w{x}$ is 15±0.15cm  

From first principles, k=F/x where F is force exerted on spring and x is displacement, k is spring constant  

Taking k as 1500 and x as 15  

F=kx=1500*15=22500N  

The partial derivative of force with respect to k yields  

$\frac {\delta F}{\delta K}= \frac {\delta}{\delta k}(kx)=(x) \frac {\delta}{\delta k}(k)=(x)(1)=x$  

Therefore, $\frac {\delta F}{\delta K}=x$=15  

The partial derivative of force with respect to x yields  

$\frac {\delta F}{\delta x}= \frac {\delta}{\delta x}(kx)=(k) \frac {\delta}{\delta x}(x)=(k)(1)=k$  

Therefore, $\frac {\delta F}{\delta x}=k$=1500  

The maximum possible error for force  

$(w_{F})_{max}= |w_{k}\frac {\delta F}{\delta K}|+| w_{x}\frac {\delta F}{\delta x}|$  

Substituting 15 for $w_{k}$, 0.15 for $w_{x}$ 15 for $\frac {\delta F}{\delta K}$ and 1500 for  $\frac {\delta F}{\delta x}$ for 1500  

$(w_{F})_{max}$=±[(15)(15)+(0.15)(1500)]= ±[225+225]= ±450  

Therefore, maximum possible error is ±450  

Uncertainty of measured force in absolute terms  

$w_{F}= \sqrt (({{w_{k} \frac {\delta F}{\delta k})}^{2} + {({w_{x} \frac {\delta F}{\delta x})}^{2})$  

Substituting 15 for $w_{k}$, 0.15 for $w_{x}$ 15 for $\frac {\delta F}{\delta K}$ and 1500 for  $\frac {\delta F}{\delta x}$ for 1500  

$w_{F}=±\sqrt {[(15)(15)]^{2}+ [(0.15)(1500)]^{2}}=±\sqrt (50625+50625)$= ±318.1981  

Uncertainty in absolute terms is ±318.1981  

Uncertainty of measured force in relative terms  

Relative uncertainty = $100 \frac {w_{F}}{F}$  

Since F is already calculated as 22500N and $w_{F}$=±318.1981  

Relative uncertainty= $100 \frac {318.1981}{22500}$=1.414214%  

Therefore, measured force in relative terms is 1.4%