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

It takes 52 seconds to complete 20 oscillations.

Explanation:

Step 1:

Given data  :

Bob’s mass, m = 40g = 0.04 Kg

Bob’s charge, q  $=4 \times 10^{-6} \mathrm{C}$

Step 2:

We know, by the law of Coulombs,

         F = q E

        F = ma [ Newton's law ]

      ma = qE

$a=\frac{q E}{m}$,  Here E is intensity of electric field    

Step 3:

Now,

      Put value of q , E and m

            $a=\frac{4 \times 10^{-6} \times 2.5 \times 10^{4}}{0.04}$

            $a=\frac{10 \times 10^{-6} \times 10^{4}}{0.04}$

            $a=\frac{10 \times 10^{-2}}{0.04}$  

            $a=250 \times 10^{-2}$

            $a=2.5 \mathrm{m} / \mathrm{s}^{2}$

Since, electrostatic force acts in a direction upwards.

So,  Effective acceleration = g - a

                                            = 10 - 2.5 = 7.5 m/s²

[Assuming acceleration by gravity, g = 10m/s²]        

Time given is now given by,

                                            $\mathrm{T}=2 \pi \sqrt{\frac{1}{a_{\mathrm{eff}}}}$

Case 1 :

           $\mathrm{T}=2 \pi \sqrt{\frac{1}{10}}$    -----> eqn ( 1 )

[Because in case of 1 efficient acceleration is g and g = 10 m/s² ]

Case 2 :

           $\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{1}{7.5}}$   -----> eqn ( 2 )

Now,

          $\frac{T}{T^{\prime}}=\frac{2 \pi \sqrt{\frac{1}{10}}}{2 \pi \sqrt{\frac{1}{7.5}}}$

          $\frac{T}{T^{\prime}}=\frac{\sqrt{\frac{1}{10}}}{\sqrt{\frac{1}{7.5}}}$

          $\frac{T}{T^{\prime}}=\sqrt{\frac{7.5}{10}}$

          $\frac{T}{T^{\prime}}=\sqrt{\frac{3}{4}}$

          $\frac{T}{T^{\prime}}=\frac{\sqrt{3}}{2}$

          $\frac{2 T}{\sqrt{3}}=T^{\prime}$

T is given 45 sec

          $T^{\prime}=\frac{2 T}{\sqrt{3}}$

          $T^{\prime}=\frac{2 \times 45}{\sqrt{3}}$

         $T^{\prime}=\frac{90}{\sqrt{3}}$

         $T^{\prime}=30 \times \sqrt{3}$

         $T^{\prime}=30 \times 1.732$    

         $T^{\prime}=51.96 \approx 52 \mathrm{sec}$

Thus, the required time is, 52 seconds.

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

It take 52 seconds fro 20 oscillations