Question

To determine acceleration due to gravity ,the time of 20 oscillations of a simple pendulum of length 100 cm was observed to be 40 s . calculate the value of ‘g’ and maximum percentage error in the measured value of ‘g’ .

PLEASE GIVE CORRECT PERCENTAGE ERROR...
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Answer 1

we know that

T = 2π √(L/g)

or by rearranging we get

g = 4π2L / T2

now, here

L = 100 cm = 1m

T = 40/20s = 2s

$.$

thus,

g = (4π2x1)/ 22 = 39.4384 / 4

or

g = 9.85 m/s2

the corresponding error equation would be

Δg/g = ΔL/L + 2(ΔT/T)

or

Δg = g.[ΔL/L + 2(ΔT/T)]

Answer 2

Answer:

T = 2π √(L/g)

or by rearranging we get

g = 4π2L / T2

now, here

L = 100 cm = 1m

T = 40/20s = 2s

..

thus,

g = (4π2x1)/ 22 = 39.4384 / 4

or

g = 9.85 m/s2

the corresponding error equation would be

Δg/g = ΔL/L + 2(ΔT/T)

or

Δg = g.[ΔL/L + 2(ΔT/T)]