The bob of a simple pendulum possesses a mass of 40 g and a positive charge of 4.0 10⁻⁶ C . It completes 20 oscillations in 45 s . A vertical electric field of magnitude 2.5 10^4 N C ⁻¹ pointing upward is subsequently switched on. Determine the time it will now require to execute 20 oscillations.

Step-by-step solution

The initial time period of the simple pendulum is given by T₁ = 2 L g . When the vertical upward electric field is switched on, the electric force acts upwards. The net downward acceleration of the bob becomes g_ eff = g - qE m . Given m = 40 10⁻³ kg , q = 4.0 10⁻⁶ C , E = 2.5 10^4 N C ⁻¹ , and taking g = 10 m s ⁻² : qE m = 4.0 10⁻⁶ 2.5 10^4 40 10⁻³ = 0.1 0.04 = 2.5 m s ⁻² g_ eff = 10 - 2.5 = 7.5 m s ⁻² The new time period is T₂ = 2 L g_ eff . The ratio of the time taken for 20 oscillations is equal to the ratio of