If T=2π√L/8
Then given that L=100.0±0.1cm(limit standard errors) and T=2.01±0.01s(limit standard error)
Calculate the value of g and its standard error.
Pls I need answer to that
It is a problem on error analysis
I guess you mean a pendulum
(Please type entire problem in the future)
let theta = A sin 2pi t/T
h = L (1 - cos theta)
for small theta that is
h = L theta^2/2
h max = (L/2)A^2
Max potential energy = m g h max
= m g (L/2)A^2
theta= A sin 2 pi t/T
v = A L (2 pi/T) cos 2 pi t/T
max V = 2 pi AL/T
so
max Ke = (1/2) m (2pi)^2 A^2 L^2/T^2
so
m g L/2 = (1/2) m (2 pi)^2 L^2/T^2
T^2 = (2 pi)^2 L/g
g = (2 pi)^2 L T^-2
so calculate your g =
then do
dg/dL = 2 pi^2 T^-2
dg/dT = 2 pi (-2)T^-3
now do dg = |dL dg/dL | + |dT dg/dT|
That is how the question is stated
To calculate the value of g and its standard error using the given equation T=2π√L/8, we can follow these steps:
Step 1: Determine the central value of L and T:
- L = 100.0 cm
- T = 2.01 s
Step 2: Calculate the central value of g:
From the given formula T = 2π√L/8, we can solve for g:
T = 2π√L/8
(2π√L/8) = T
√L = (8T)/(2π)
L = [(8T)/(2π)]^2
L = (16T^2)/(π^2)
Plugging in the central values of L and T:
L = (16(2.01)^2)/(π^2)
L ≈ 8.129 m/s^2
Therefore, the central value of g is approximately 8.129 m/s^2.
Step 3: Calculate the standard error of g:
We have the following standard errors:
- Standard error of L: ΔL = 0.1 cm
- Standard error of T: ΔT = 0.01 s
Using error propagation, the formula to calculate the standard error of g is:
Δg = √[(∂g/∂L)^2(ΔL)^2 + (∂g/∂T)^2(ΔT)^2]
Taking the derivatives:
∂g/∂L = ∂(√L)/∂L = 1/(2√L)
∂g/∂T = ∂[(16T^2)/(π^2)]/∂T = 32T/(π^2)
Plugging in the values:
Δg = √[(1/(2√L))^2(ΔL)^2 + (32T/(π^2))^2(ΔT)^2]
Δg = √[(1/(2√8.129))^2(0.1)^2 + (32(2.01)/(π^2))^2(0.01)^2]
Calculating Δg:
Δg ≈ 0.172 m/s^2
Therefore, the standard error of g is approximately 0.172 m/s^2.
Step 4: Summarizing the results:
The value of g is approximately 8.129 m/s^2 with a standard error of 0.172 m/s^2.