A simple pendulum of length l and a tiny metal bob of mass m is placed between t

Question

A simple pendulum of length l and a tiny metal bob of mass m is placed between two large conducting parallel plates as shown. How will the natural frequency of oscillations change if the bob is given charge +e, when the upper plate is : (a)Positively charged (b)Negatively charged. Assume m=1g and that the period of oscillations in absence of charges is T1= (pi/5)second. Once the charges are introduced the period is measured to be T2=(T1/2)second. Find (c)The length of the pendulum l.(d)The force exerted by the E-field of the plates on a bob. (e)The period of oscillations if the charge of the bob is -e.

length l and a tiny metal bob of mass m is placed between parallel plates as shown. How will thye natural frequency of the bob is given cahrge + e, when the upper plate is:

Explanation / Answer

a)

As the frequency is directly proportional to the square root of the net vertical acceleration,

So, when the upper plate is +vely charged, thus the bob will be acted upon by two vertical forces, the gravitational force downwards and electric force also facing downwards,

So, frequency increases <--------------answer

b)

just the opposite happens, So frequency decreases <-------answer

c)

f2/f1 = sqrt(a2/a1) = 2

so, a2/a1 = 4

so, a = 3g

So, T = pi/5 = 2*pi*sqrt(l/9.8)

so, l = 0.098 m <-------------answer(length of the pendulum)

d)

force exerted = E*e = 3mg = 3*1*9.8 = 29.4 N <-----------answer

e)

period = 2*pi*sqrt(0.098/4*9.8) = 3.08 s <-----------answer

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