A simple pendulum with a bob of mass 40 g an | vedclass

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(A) Given: Mass $m = 40 \ g = 0.04 \ kg$,charge $q = 2 	imes 10^{-6} \ C$,$g = 10 \ ms^{-2}$,electric field $E = 4.2 	imes 10^4 \ NC^{-1}$.First,cal

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(A) Given: Mass $m = 40 \ g = 0.04 \ kg$,charge $q = 2 	imes 10^{-6} \ C$,$g = 10 \ ms^{-2}$,electric field $E = 4.2 	imes 10^4 \ NC^{-1}$.First,calculate the acceleration due to the electric field:$a = \frac{qE}{m} = \frac{2 	imes 10^{-6} 	imes 4.2 	imes 10^4}{0.04} = \frac{0.084}{0.04} = 2.1 \ ms^{-2}$.Since the electric field is downward and the charge is positive,the force is downward,so the effective acceleration $g_e = g + a = 10 + 2.1 = 12.1 \ ms^{-2}$.The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.In the absence of the field,$T = \frac{44}{20} = 2.2 \ s$.Thus,$2.2 = 2\pi \sqrt{\frac{l}{10}} \Rightarrow \sqrt{l} = \frac{2.2 \sqrt{10}}{2\pi}$.In the presence of the field,$T' = 2\pi \sqrt{\frac{l}{12.1}} = 2\pi \frac{\sqrt{l}}{\sqrt{12.1}}$.Substituting $\sqrt{l}$,we get $T' = 2\pi \left( \frac{2.2 \sqrt{10}}{2\pi \sqrt{12.1}} ight) = 2.2 \sqrt{\frac{10}{12.1}} = 2.2 	imes \frac{\sqrt{10}}{1.1 \sqrt{10}} = 2.2 	imes \frac{1}{1.1} = 2 \ s$.The time taken for $15$ oscillations is $t = 15 	imes T' = 15 	imes 2 = 30 \ s$.

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