A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :

Question

A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is : (A) 2.40 % (B) 5.40 % (C) 4.40 % (D) 3.40 %

Tardigrade Admin · Accepted Answer

T=2π√(ℓ/g) g=(4π2ℓ/T2) (Δ g/g)=(Δℓ/ℓ)+(2Δ T/T) =(0.1/05)+(2×1/50) (Δ g/g)=4.4 %

Q. A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is $25.0 c m$ and a stop watch with $1 s$ resolution measures the time taken for $40$ oscillations to be $50 s$ . The accuracy in $g$ is :

$2.40%$

$5.40%$

$4.40%$

$3.40%$

$T = 2 π \frac{ℓ}{g}$
$g = \frac{4 π ^{2} ℓ}{T ^{2}}$
$\frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + \frac{2Δ T}{T}$
$= \frac{0.1}{05} + \frac{2 \times 1}{50}$
$\frac{Δ g}{g} = 4.4%$

Q. A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50s. The accuracy in g is :

2.40%

5.40%

4.40%

3.40%