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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?
Solution
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
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