Answer:
Maximum possible error is ±450
Uncertainty in absolute terms is ±318.1981
Relative terms is 1.4%
Explanation:
Spring constant, [tex]k±w_{k}[/tex] is 1500±15 N/cm and displacement, [tex]x±w{x}[/tex] 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
[tex]\frac {\delta F}{\delta K}= \frac {\delta}{\delta k}(kx)=(x) \frac {\delta}{\delta k}(k)=(x)(1)=x[/tex]
Therefore, [tex]\frac {\delta F}{\delta K}=x[/tex]=15
The partial derivative of force with respect to x yields
[tex]\frac {\delta F}{\delta x}= \frac {\delta}{\delta x}(kx)=(k) \frac {\delta}{\delta x}(x)=(k)(1)=k[/tex]
Therefore, [tex]\frac {\delta F}{\delta x}=k[/tex]=1500
The maximum possible error for force
[tex](w_{F})_{max}= |w_{k}\frac {\delta F}{\delta K}|+| w_{x}\frac {\delta F}{\delta x}|[/tex]
Substituting 15 for [tex]w_{k}[/tex], 0.15 for [tex]w_{x}[/tex] 15 for [tex]\frac {\delta F}{\delta K}[/tex] and 1500 for [tex]\frac {\delta F}{\delta x}[/tex] for 1500
[tex](w_{F})_{max}[/tex]=±[(15)(15)+(0.15)(1500)]= ±[225+225]= ±450
Therefore, maximum possible error is ±450
Uncertainty of measured force in absolute terms
[tex]w_{F}= \sqrt (({{w_{k} \frac {\delta F}{\delta k})}^{2} + {({w_{x} \frac {\delta F}{\delta x})}^{2})[/tex]
Substituting 15 for [tex]w_{k}[/tex], 0.15 for [tex]w_{x}[/tex] 15 for [tex]\frac {\delta F}{\delta K}[/tex] and 1500 for [tex]\frac {\delta F}{\delta x}[/tex] for 1500
[tex]w_{F}=±\sqrt {[(15)(15)]^{2}+ [(0.15)(1500)]^{2}}=±\sqrt (50625+50625) [/tex]= ±318.1981
Uncertainty in absolute terms is ±318.1981
Uncertainty of measured force in relative terms
Relative uncertainty = [tex]100 \frac {w_{F}}{F}[/tex]
Since F is already calculated as 22500N and [tex]w_{F}[/tex]=±318.1981
Relative uncertainty= [tex]100 \frac {318.1981}{22500}[/tex]=1.414214%
Therefore, measured force in relative terms is 1.4%
