Question

9. The pened of oscillation of a simple pendulum siven by T
where
about 100
by
known to have I mm accuracy. The period is about
stop watch of least count 1. The percentage error in
(A) 019
(B) 19
the time of 100 oscillations
s
DIO​

Answer 1

T =2π√( L/g )

t/n = 2π√L/g

t= 4π²Ln²/g

g = 4π²L/t²

here n is no of oscillations

so,

∆g/g = ∆L/L + 2∆t/t

given ,

L = 100 cm

∆L = 1mm = 0.1 cm

t = 200 sec

∆t =0.1 sec

now,

∆g/g = 0.1/100 + 2× 0.1/200

percentage error % of g = ∆g/g ×100

=0.1/100 ×100 + 0.1×100/100

=0.1 + 0.1= 0.2 %