Answer:
g = 9.82
standard error = 0.0063
Step-by-step explanation:
To solve this problem, we can use the given equation T = 2 phi root L / g, where T is the period of oscillation, L is the length of the pendulum, and g is the acceleration due to gravity. We are given the values of T and L, and we are asked to calculate the value of g and its standard error.
First, we can calculate the value of g by substituting the given values for T and L into the equation:
T = 2 phi root L / g
2.01 = 2 phi root 100.0 / g
g = 2 phi root 100.0 / 2.01
g = 9.82
Next, we can calculate the standard error of g by propagating the uncertainties in T and L through the equation. To do this, we can use the following formula:
standard error of g = sqrt(((derivative of g with respect to T) * standard error of T)^2 + ((derivative of g with respect to L) * standard error of L)^2)
The derivative of g with respect to T is -2 phi root L / T^2, and the derivative of g with respect to L is -2 phi / (g * L^(1/2)). Substituting these values into the formula, we get:
standard error of g = sqrt(((-2 phi root L / T^2) * 0.01)^2 + ((-2 phi / (g * L^(1/2))) * 0.1)^2)
standard error of g = sqrt(((-2 * 3.14 * sqrt(100.0) / 2.01^2) * 0.01)^2 + ((-2 * 3.14 / (9.82 * sqrt(100.0))) * 0.1)^2)
standard error of g = sqrt(((-0.63 * 0.01)^2 + (-0.062 * 0.1)^2))
standard error of g = sqrt((-0.0063^2 + -0.0062^2))
standard error of g = sqrt(0.00003969 + 0.00003844)
standard error of g = 0.0063
Thus, the value of g is 9.82, and its standard error is 0.0063.
I hope this helps! Let me know if you have any other questions or need further assistance.