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

Maximum possible error is ±450

Uncertainty in absolute terms is ±318.1981

Relative terms is 1.4%

Step-by-step 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%