The length of a simple pendulums which is about 0.5m can be measured to within 1mm what accuracy is required in measurement of time of 1000 oscillations. If error in L and T are to produce equal percentage in the calculated value of g?
To find the accuracy required in the measurement of time of 1000 oscillations, considering the equal percentage error between length (L) and time (T), we need to use the formula for the period (T) of a simple pendulum:
T = 2π √(L/g)
Here, L represents the length of the pendulum, g represents the acceleration due to gravity, and T represents the period of oscillation.
Let's proceed step by step:
1. Calculate the relative error in the length of the pendulum (ΔL/L):
relative error in length (ΔL/L) = absolute error in length (ΔL) / length of the pendulum (L)
Given that the length of the pendulum can be measured within 1 mm (ΔL = 0.001 m), and the length is 0.5 m:
ΔL/L = 0.001 m / 0.5 m = 0.002
2. Determine the relative error in the period (ΔT/T):
relative error in time (ΔT/T) = absolute error in time (ΔT) / time period (T)
We need to find the value of ΔT/T such that it produces an equal percentage error as ΔL/L when calculating the value of 'g.'
Since the equation for the pendulum period is T = 2π √(L/g), we can take the natural logarithm of both sides:
ln(T) = ln(2π) + (1/2) ln(L/g)
Now, let's differentiate both sides with respect to 'T':
(1/T) dT = (1/2) [ (1/L) dL - (1/g) dg ]
Simplifying the equation:
dT/T = (dL/L - dg/g) / 2
By applying the concept of equal percentage error, we set dL/L = dg/g:
dT/T = 0.002 / 2 = 0.001
Therefore, we found that the relative error in the period of oscillation, ΔT/T, is 0.001.
3. Calculate the absolute error in the period (ΔT) by multiplying the relative error (ΔT/T) with the value of the period (T):
ΔT = ΔT/T * T = 0.001 * T
Here, T represents the actual period of oscillation, which we need to find.
4. Solve for the actual period (T) by rearranging the equation for the period of a simple pendulum:
T = 2π √(L/g)
Given that the length of the pendulum is 0.5 m and acceleration due to gravity is approximately 9.8 m/s², we substitute these values into the equation:
T = 2π √(0.5/9.8)
Calculate the actual period (T) using this equation.
5. Finally, substitute the obtained T value into the absolute error formula to find the accuracy required in the measurement of time:
ΔT = 0.001 * T
This will give you the accuracy required in the measurement of time of 1000 oscillations when the error in length (L) and time (T) produce an equal percentage error in the calculated value of 'g'.