The time period of oscillation of a simple pendulum is T = 2π√(l/g). Measured value of l is 10 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 50 s using a wrist watch of Is resolution. What is the accuracy in the determination of g ?

Question

The time period of oscillation of a simple pendulum is T = 2π√(l/g). Measured value of l is 10 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 50 s using a wrist watch of Is resolution. What is the accuracy in the determination of g ? (A) 2 % (B) 3 % (C) 4 % (D) 5 %

Tardigrade Admin · Accepted Answer

Here, T=2π√(l/g) ∴ Relative error in g is (Δ g/g)=(Δ l/l)+(2Δ T/T). Here, T=(t/n), and Δ T=(Δ t/n) ∴ (Δ T/T)=(Δ t/t). The errors in both l and t are least count errors. ∴ (Δ g/g)=(0.1/10)+2((1/50)) =0.01+0.04=0.05 The percentage error in g is (Δ g/g)× 100=(Δ l/l)× 100+2((Δ T/T))× 100 =[(Δ l/l)+2((Δ T/T))]× 100 =0.05× 100=5 %

Q. The time period of oscillation of a simple pendulum is $T = 2 π \frac{l}{g}$ . Measured value of $l$ is $10 c m$ known to $1$ mm accuracy and time for $100$ oscillations of the pendulum is found to be $50 s$ using a wrist watch of Is resolution. What is the accuracy in the determination of $g$ ?

$2%$

$3%$

$4%$

$5%$

Q. The time period of oscillation of a simple pendulum is T=2πgl​​. Measured value of l is 10cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 50s using a wrist watch of Is resolution. What is the accuracy in the determination of g ?

2%

3%

4%

5%

$2%$

$3%$

$4%$

$5%$