The period of oscillation of a simple pendulum is T =2 π √( 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 80 s using a wrist watch of 1 s accuracy. What will be the accuracy (in %) in the determination of g?

Question

The period of oscillation of a simple pendulum is T =2 π √( 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 80 s using a wrist watch of 1 s accuracy. What will be the accuracy (in %) in the determination of g? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Time period, T =2 π √( L / g ) ∴ g =4 π2 ( L / T 2) ∴ (Δ g / g ) × 100 =(Δ L / L ) × 100+2 (Δ T / T ) × 100 =(0.1/20.0) × 100+2 × (1/80) × 100 =(10/20)+(200/80)=(1/2)+(20/8) =3 %

Q. The period of oscillation of a simple pendulum is T=2πgL​​. Measured value of L is 20.0cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 80s using a wrist watch of 1s accuracy. What will be the accuracy (in %) in the determination of g?

Time period, T=2πgL​​ ∴g=4π2T2L​ ∴gΔg​×100=LΔL​×100+2TΔT​×100 =20.00.1​×100+2×801​×100 =2010​+80200​=21​+820​ =3%

Time period, $T = 2 π \frac{L}{g}$
$∴ g = 4 π^{2} \frac{L}{T ^{2}}$
$∴ \frac{Δ g}{g} \times 100 = \frac{Δ L}{L} \times 100 + 2 \frac{Δ T}{T} \times 100$
$= \frac{0.1}{20.0} \times 100 + 2 \times \frac{1}{80} \times 100$
$= \frac{10}{20} + \frac{200}{80} = \frac{1}{2} + \frac{20}{8}$
$= 3%$