Question

The period of oscillation of a simple pendulum is given by where l is about 100 cm and is known to have 1mm accuracy. The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is

Answer 1

Answer:

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 %

YW! :)