Question:

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10cm to 8cm in 40s. Assuming that Stokes law is valid, and the ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3, the time in which the amplitude of this pendulum will reduce from 10cm to 5cm in carbon dioxide will be close to (ln5=1.601, ln2=0.693).

142s

208s

231s

161s

The correct option is B. 161s

In air, 10cm → 8cm in 40s
In CO2, 10cm → 5cm in ?
ηair/ηCO2 = 1.3

We know that the amplitude of damped oscillation is given by,
A = A0e-bt/2m

→ 8 = 10e(-bair × 40/2m) and 5 = 10e(-bCO2 × t/2m)

→ ln(⅘) = -bair × 40/2m and ln(½) = -bCO2 × t/2m

Therefore, t = (bair/bCO2) × 40 × (ln(½)/ln(⅘))

= 1.3 × 40 × (0.693/1.601) ≈ 161s