The length l of a simple pendulum is about 0.5m, and can be measured within 1mm. what accuracy is required in the measurement of periodic time T of 100 oscillations if the error in l and T are to produce equal percentage in the calculation of acceleration due to gravity (g)
The length L measurement relative error is 1/500 = 0.2%
g = L*(200 pi)^2/T^2
The relative error in T should be 0.1% if its contribution to total error equals the relative error in L.
thanks!!
thanks alot...it was an easy method :)
To find the required accuracy in the measurement of the periodic time (T) of 100 oscillations, such that the error in length (l) and T produce an equal percentage error in the calculation of acceleration due to gravity (g), we need to consider the equation relating the period of a simple pendulum with its length and gravitational acceleration.
The equation for the period of a simple pendulum is given by:
T = 2π√(l/g)
where T is the period, l is the length of the pendulum, and g is the acceleration due to gravity.
Let's denote the errors in the measurement of l and T as Δl and ΔT, respectively.
The percentage error in the calculation of g due to the error in l can be given as:
(Δg/g)l = (Δl/l)
Similarly, the percentage error in the calculation of g due to the error in T can be given as:
(Δg/g)T = (ΔT/T)
Since we want the errors to produce equal percentage errors, we can equate the two expressions:
(Δl/l) = (ΔT/T)
Now, let's substitute the given values:
Length (l) = 0.5 m
Error in length (Δl) = 1 mm = 0.001 m
Period (T) = ?
We can rearrange the equation to solve for T:
T = 2π√(l/g)
Squaring both sides:
T^2 = 4π^2(l/g)
Solving for g:
g = 4π^2(l/T^2)
Now, let's differentiate both sides with respect to l and T:
dg/dl = 4π^2(1/T^2)
dg/dT = -8π^2(l/T^3)
Let's consider the absolute values of the derivatives:
|dg/dl| = 4π^2/T^2
|dg/dT| = 8π^2l/T^3
Since we want the errors to produce equal percentage errors, we compare the absolute values:
(Δg/g)l = (Δl/l) = |dg/dl| = 4π^2/T^2
(Δg/g)T = (ΔT/T) = |dg/dT| = 8π^2l/T^3
Now, we equate the two expressions:
4π^2/T^2 = 8π^2l/T^3
Simplifying:
T = 2l
Therefore, the required accuracy in the measurement of T would be twice the accuracy of the measurement of l.
In our case, the error in l is 0.001 m, so the required accuracy in the measurement of T would be 2 * 0.001 m = 0.002 m.