Question

For a simple pendulum of length 50 cm ± 0.1, the time period for 100 oscillations is found to be 50 s when observed using a stop watch having resolution of 1 s. Find the percentage error in determination of g.​

Answer 1

Answer:

given = length = 50cm +0.1

            T = 50s , no. of oscillation = 100

to find = %  error in determination of g

solution,

T =  2a $\sqrt{\frac{t}{g} }$

T²= 4π2L/ g

g = 4π² L/g

error is determination of g.

Δg/g = ΔK/L + 2(ΔT/T)

T = t/h = ΔT = 1s

ΔT/T  =  Δt / n

Δg / g = 0.1 / 50 +  2(1/50 ) =  0.42%

hence the answer is 0.42%