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The time period of simple pendulum is t=2π.what is the accuracy in determination of g if 10cm length is known to 1mm accuracy and 0.5 s time period is measured from time of 100 oscillations with the watch of 1sec resolution?
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Calculation of Accuracy in Determination of g
Given Data:
- Length of the pendulum (l) = 10 cm
- Accuracy in the measurement of length (Δl) = 1 mm
- Time period of the pendulum (t) = 0.5 s
- Number of oscillations (n) = 100
- Resolution of the watch (Δt) = 1 s
Calculation of g:
The time period of a simple pendulum is given by the formula:
t = 2π * sqrt(l/g)
where g is the acceleration due to gravity.
On rearranging the above formula, we get:
g = (4π^2 * l) / t^2
Substituting the given values, we get:
g = (4π^2 * 0.1) / (0.5)^2 = 39.478 m/s^2
Calculation of Accuracy in Determination of g:
The percentage error in the measurement of length is given by:
Δl/l * 100%
Substituting the given values, we get:
Δl/l * 100% = (1/10) * 100% = 10%
Similarly, the percentage error in the measurement of time is given by:
Δt/t * 100% = (1/500) * 100% = 0.2%
The total percentage error in the determination of g is given by:
Δg/g * 100% = (Δl/l + 2Δt/t) * 100%
Substituting the given values, we get:
Δg/g * 100% = (10% + 2*0.2%) * 100% = 10.4%
Therefore, the accuracy in the determination of g is 10.4%.
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