The period of oscillation of a simple pendulum is T=2π√(L/g). Measured of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g?
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Answer:
T= Time period of one oscillation
t= Time period of 100 oscillation
For 100 oscillation, t=100T
Take ln and differentiate.
t
Δt
=
T
ΔT
=
90
2
.....(1)
Time period of one oscillation
T=2π
g
L
g=4π
2
T
2
L
Take ln on both side and differentiate
g
Δg
=±(
L
ΔL
+2
T
ΔT
)
g
Δg
=±(
20
0.1
+
90
2
)==±2.7%
Hence, error in gravity is ±2.7%
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