Question

The period of oscillation of a simple pendulum is
$t = 2\pi \sqrt{ \frac{l}{g} } .$
Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 seconds using a wrist watch of 1 s revolution. What is the accuracy in the determination of g?

Answer 1

HELLO DEAR,

given that:-

∆l = 1mm = 0.1cm
l = 20cm
∆t = 1
t = 90

t = 2π√l/g

on squaring both side

we get,

t² = 4π² l/g

=> ∆g/g × 100 = ∆l/l× 100 + 2×∆t/t × 100

=> ∆g/g × 100 = 0.1/20 × 100 + 2 × 1/90 × 100

=> 0.5 + 2.2 = 2.7

=> ∆g/g = 2.7 = similarly 3

I HOPE ITS HELP YOU DEAR ,THANKS

Answer 2

Hello friend....!!

According to the given question :

We know that the time period of the simple pendulum is

T = 2π √ L / g --------( 1 )

Now when we square on both sides we get :

T^2 =( 4π^2 ) L / g

From here,

G = 4π^2 L / T^2

Now to find the accuracy in the determination of g :

Δg / g = { ΔL / L + 2 ( ΔT/ T) }

Implies,

Here it is in the form of least count error

Δg / g = { 0.1 / 20.0 + 2 ( 1 / 90 ) }

Δg / g = 0.027

Also,

Δg / g = 0.027 x 100%

Implies,

Δg / g = 2.7% .

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Hope it helps...!!!