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A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the pendulum, g is gravity, andT is the period of pendulum.Twenty measurements of T give a mean of 1.823 seconds and a standard deviation of 0.0671 s. The device used to measure time has a resolution of 0.02 s. The pendulum length is measured once to be 0.823 m (with a scale having a resolution of 0.001 m). Determine the value of g and its uncertainty (assume 90% confidence where necessary). You may use any method of uncertainty propagation that we covered in class.

Sagot :

Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

          T = [tex]2\pi \sqrt{ \frac{L}{g} }[/tex]

          T² =  [tex]4\pi ^2 \frac{L}{g}[/tex]4pi2 L / g

          g = [tex]4\pi ^2 \frac{L}{T^2}[/tex]

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

let's calculate the magnitude

           g = [tex]4\pi ^2 \frac{0.823}{1.823^2}[/tex]4 pi2 0.823 / 1.823 2

            g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

for the period

             T = t / n

             ΔT = [tex]\frac{dT}{dt}[/tex] Δt + [tex]\frac{dT}{dn}[/tex] ΔDn

In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently

              ΔT = Δt / n

              ΔT = Δt

now let's look for the uncertainty of g

             Δg = [tex]\frac{dg}{dL}[/tex] ΔL + [tex]\frac{dg}{dT}[/tex]  ΔT

             Δg = [tex]4\pi ^2 \frac{1}{T2}[/tex]   ΔL + 4π²L  (-2  T⁻³) ΔT

           

a more manageable way is with the relative error

             [tex]\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}[/tex]

we substitute

              Δg = g ( \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}DL / L + ½ Dt / T)

the error in time give us the stanndard deviation  

let's calculate

               Δg = 9.7766 ([tex]\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}[/tex])

               Δg = 9.7766 (0.001215 + 0.0184)

               Δg = 0.19 m / s²

the absolute uncertainty must be true to a significant figure

                Δg = 0.2 m / s2

therefore the correct result is

               g ±Δg = (9.8 ± 0.2) m / s²

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