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.