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?

(a) ¼

(b) ½

(c) 1

(d) 2

(e) 4

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?

(a) *L*

(b) √6 *L*

(c) *L/*√6

(d) *L*/6

(e) 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 *L*_{1}, we have

*T = *2π√(*L*_{1}/g_{1}) where g_{1} = g/6.

Comparing the above equations, we have *L*_{1} = *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?

(a) 2π√(*L*/3*g*)

(b) 2π√(*L*/5*g*)

(c) 4π√(*L*/*g*)

(d) 2π√(*L*/*g*)

(e) 4π√(*L*/5*g*)

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√(4*L*/5*g*) = 4π√(*L*/5*g*)

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 + *q*^{2}/4πε_{0}*L*^{2})]^{1/2}

(b) 2π [*L*/(g – *q*^{2}/4πε_{0}*L*^{2})]^{1/2}

(c) 2π [*L*/g)]^{1/2}

(d) 2π [*L*/(g + *q*^{2}/4πε_{0}*L*^{2}*m*)]^{1/2}

(e) 2π [*L*/(g – *q*^{2}/4πε_{0}*L*^{2}*m*)]^{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 *q*^{2}/4πε_{0}*L*^{2} due to the electrostatic repulsion. Thus the apparent weight of the pendulum bob is (*m*g+ *q*^{2}/4πε_{0}*L*^{2}). The acceleration due to gravity ‘g’ appearing in the above expression for the period has therefore to be replaced by g_{1} = (*m*g+ *q*^{2}/4πε_{0}*L*^{2})/*m* = (g + *q*^{2}/4πε_{0}*L*^{2}*m*).

The period of oscillation *T*_{1}* *of the pendulum is therefore given by

*T*_{1} = 2π [*L*/(g + *q*^{2}/4πε_{0}*L*^{2}*m*)]^{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 + *q*^{2}/4πε_{0}*L*^{2}*m*)]^{1/2}

(b) 2π [*L*/(g – *q*^{2}/4πε_{0}*L*^{2}*m*)]^{1/2}

(c) 2π [*L*/(g + *q*^{2}/4πε_{0}*L*^{2})]^{1/2}

(d) 2π [*L*/(g – *q*^{2}/4πε_{0}*L*^{2})]^{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 g_{2} = (*m*g – *q*^{2}/4πε_{0}*L*^{2})/*m* = (g – *q*^{2}/4πε_{0}*L*^{2}*m*).

The period of oscillation *T*_{2}* *of the pendulum is therefore given by

*T*_{2} = 2π [*L*/(g – *q*^{2}/4πε_{0}*L*^{2}*m*)]^{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)

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