Question

The bob of a simple pendulum has a mass of 40 g and a positive charge of 4.0 × 10⁻⁶ C. It makes 20 oscillations in 45 s. A vertical electric field pointing upward and of magnitude 2.5 × 10⁴ NC⁻¹ is switched on. How much time will it now take to complete 20 oscillations? 

Answer 1

Given:
Charge of the particle, q = 4.0 × 10⁻⁶ C
Electrical field intensity, E = 2.5 × 10⁴ NC⁻¹
Mass of the particle, m = 40 g = 0.04 kg
When no electrical field is applied,
time period,

\[T_1  = 2\pi\sqrt{\frac{l}{g}}\] 

When upward electrical field is applied,
time period,

\[T_2  = 2\pi\sqrt{\frac{l}{g - a}}\],

where a is the upward acceleration due to electrical field, which is given by

\[a = \frac{qE}{m}\] 

\[ \Rightarrow a = \frac{4 \times {10}^{- 6} \times 2 . 5 \times {10}^4}{40 \times {10}^{- 3}}\] 

\[               = 2 . 5  \text{m/ s}^2 \]

\[\therefore \frac{T_2}{T_1} = \sqrt{\frac{g}{g - a}}\] 

\[ \Rightarrow \frac{T_2}{T_1} = \sqrt{\frac{9 . 8}{9 . 8 - 2 . 5}}\] 

\[ \Rightarrow  T_2  = 45 \times \sqrt{\frac{9 . 8}{7 . 3}} = 52  \] s