Today we will discuss a few multiple choice practice questions involving simple pendulum. Electronic and digital clocks have replaced the old mechanical pendulum clocks; but the pendulum is important in the study of oscillations. The simple pendulum is a really simple system when we consider its oscillations with small amplitude. Here are the questions:
(1) By the term ‘seconds pendulum’ we mean a pendulum of period two seconds. What is the frequency (in Hz) with which the kinetic energy oscillates in the case of a seconds pendulum?
During each complete oscillation of the pendulum the kinetic energy passes through two maxima (when the pendulum bob is in the mean position) and two minima (which is zero, in the extreme positions). The frequency of oscillation of the kinetic energy is therefore twice the frequency of oscillation of the pendulum. Since the frequency of oscillation of the seconds pendulum is ½ (since frequency, f = 1/T where T is the period). Therefore, the frequency (in Hz) with which the kinetic energy oscillates in the case of a seconds pendulum.is 1. [Option (c)].
(2) The gravitational acceleration on the moon’s surface is approximately g/6 where ‘g’ is the gravitational acceleration on the earth’s surface. A simple pendulum of length L has a period T. on the surface of the earth. What should be the approximate length of this pendulum so as to have the same period T on the surface of the moon?
(b) √6 L
The period of oscillation T of the simple pendulum is given by
T = 2π√(L/g)
If the corresponding length of the pendulum (to give the same period T) on the moon is L1, we have
T = 2π√(L1/g1) where g1 = g/6.
Comparing the above equations, we have L1 = L/6.
(3) A simple pendulum arranged inside an elevator has a period T when the elevator is at rest. When the elevator is moving down with deceleration g/4 where ‘g’ is the magnitude of the acceleration due to gravity, what will be the period of oscillation of the pendulum inside the elevator?
If the elevator moves down with an acceleration ‘a’, an object of mass ‘m’ inside the elevator will have a weight m(g – a).
[You might have felt a reduction in your weight when you accelerate downwards in a swing].
Since the elevator is decelerating while moving down, the weight of the bob of the pendulum will be m[g – (– a)] = m(g+a).
Therefore, in the expression 2π√(L/g) for the period of the pendulum, g is to be replaced by (g+a) when the elevator moves down with a deceleration of magnitude a.
Since a = g/4, the period of the pendulum is 2π√[L/(g+ g/4)] = 2π√(4L/5g) = 4π√(L/5g)
The following questions are meant for AP Physics C aspirants even though AP Physics B aspirants also can answer them.
(4) The spherical metallic bob A (fig.) of a simple pendulum has mass m and is hanging vertically down from an identical fixed bob B by means of a string of length L. Both bobs carry the same positive charge q and there are no electric fields other than those produced by the charged bobs. If the acceleration due to gravity is g, what is the period of the pendulum?
(a) 2π [L/(g + q2/4πε0L2)]1/2
(b) 2π [L/(g – q2/4πε0L2)]1/2
(c) 2π [L/g)]1/2
(d) 2π [L/(g + q2/4πε0L2m)]1/2
(e) 2π [L/(g – q2/4πε0L2m)]1/2
If the bobs are uncharged the period T of the pendulum is given by
T = 2π√(L/g) as usual.
Since the bobs are charged, there is an extra downward force q2/4πε0L2 due to the electrostatic repulsion. Thus the apparent weight of the pendulum bob is (mg+ q2/4πε0L2). The acceleration due to gravity ‘g’ appearing in the above expression for the period has therefore to be replaced by g1 = (mg+ q2/4πε0L2)/m = (g + q2/4πε0L2m).
The period of oscillation T1 of the pendulum is therefore given by
T1 = 2π [L/(g + q2/4πε0L2m)]1/2
(5) If the bobs in question number 3 carry equal but opposite charges and the pendulum oscillates, what will be the period of the pendulum?
(a) 2π [L/(g + q2/4πε0L2m)]1/2
(b) 2π [L/(g – q2/4πε0L2m)]1/2
(c) 2π [L/(g + q2/4πε0L2)]1/2
(d) 2π [L/(g – q2/4πε0L2)]1/2
(e) 2π [L/g)]1/2
In this case there is an electrostatic attractive force between the bobs and the acceleration due to gravity is to be replaced by g2 = (mg – q2/4πε0L2)/m = (g – q2/4πε0L2m).
The period of oscillation T2 of the pendulum is therefore given by
T2 = 2π [L/(g – q2/4πε0L2m)]1/2, as given in option (d).
Now, suppose a simple pendulum is arranged as usual using a tall stand in an electric field E directed vertically downwards. If the bob of the pendulum carries a charge +q, the period of oscillation of the pendulum will be decreased since the apparent weight of the pendulum bob will be mg+ qE so that in the usual expression 2π√(L/g) for the period of the pendulum, you have to replace g with g+ (qE/m).
If the bob has a negative charge (–q), the period will be increased since you have to replace g with g– (qE/m). This is also the case with a positively charged bob and an upward electric field. In these cases the gravitational force on the bob is usually given to be greater than the upward electric force so as to keep the string of the pendulum taut and the pendulum oscillates in the usual manner.
[If the upward electric force is greater than the gravitational force, the pendulum still oscillates, but with the bob pointing upwards! In that case you will have to replace g with [(qE/m) –g] in the usual expression for the period].
You will often encounter questions on simple pendulum with charged bob arranged in vertical electric fields. You will find a useful post here on oscillations (simple harmonic motion)