AP Physics B & C – Oscillations & Gravitation – Multiple Choice Practice Questions on Simple Pendulum

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) 6L

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?

(a) 2π√(L/3g)

(b) 2π√(L/5g)

(c) 4π√(L/g)

(d) 2π√(L/g)

(e) 4π√(L/5g)

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)

AP Physics C – Applications of Ampere’s Circuital Law – Magnetic Field due to Straight Infinitely Long Thick Current Carrying Cylinders and Pipes

In the post dated 1^st February 2008 the equations to be noted in connection with magnetic fields were given. [To access that post click here]. Ampere’s circuital law was briefly mentioned there.

Ampere’s circuital law states that the line integral of magnetic flux density B over any closed curve is equal to µ₀ times the total current I passing through the surface enclosed by the closed curve. This is stated mathematically as

∫B. dℓ = µ₀I (The integration is over the closed path)

[Ampere’s circuital law as modified by Maxwell to accommodate the displacement current flowing through dielectrics and free space is

∫B. dℓ = µ₀ [I+ ε₀ (dφ_E/dt)], where ε₀ (dφ_E/dt) is the displacement current resulting from the rate of change of electric flux φ_E. ε₀ is the permittivity of free space].

In this post I want to make you realize the importance of Ampere’s circuital law as a simple tool for evaluating the magnetic field in situations exhibiting symmetry in such a manner that the magnitude of the magnetic field is the same along the path chosen for the line integral.

You might have used Ampere’s circuital law to derive the expression, B = µ₀µ_rnI for the magnitude of the magnetic flux density inside very long solenoids and toroids. [Here µ₀ is the permeability of free space, µ_r is the relative permeability of the medium filling the interior of the solenoid (or toroid), n is the number of turns per metre and I is the current].

You might have certainly used Ampere’s circuital law to derive the expression, B = μ₀I/2πr for the magnitude of the magnetic flux density at a point P distant ‘r’ from an infinitely long straight conductor carrying a steady current I. The above expression shows that the magnetic field is inversely proportional to the distance of the point P from the conductor. Normally you won’t worry about the variation of the magnetic field inside the conductor since the conductor is assumed to be thin. Now, suppose the conductor is thick. Now you have enough space to move about inside the body of the conductor to explore the variation of the magnetic field!

Let us use Ampere’s circuital law for finding the magnetic field due to a thick cylindrical conductor of radius R. A cross section of the thick conductor is shown in the following figure.

The conductor is straight and infinitely long, with its length normal to the plane of the figure. The current I is supposed to flow into the plane of the figure, away from the reader.

Consider a point P at distance r from the centre of the conductor such that r < R so that the point P is within the conductor. This is a situation exhibiting symmetry and the magnitude B of the magnetic flux density at all points at the same distance r from the centre must be the same. Applying Ampere’s circuital law, we equate the line integral of the magnetic field vector B over a circular path (red dotted circle) of radius r to the product of µ₀ and the appropriate current:

∫B. dℓ = µ₀I₁ ………….(i)

where I₁ is the current passing (piercing) through the surface enclosed by the circular path of radius r. Note that I₁ is the current carried by the cylindrical portion of the conductor having radius ‘r’so that

I₁ = I(πr²/πR²) = Ir²/R²

[The total current I passes through the total cross section area πR². So the current I₁ passing through the cross section area πr² is I(πr²/πR²) = Ir²/R²].

Substituting for I₁ in equation (i) and remembering that the direction of the magnetic field B is along the path of integration everywhere, we have

2πrB = µ₀Ir²/R²

Therefore, B = μ₀Ir/2πR²

This shows that the magnetic field at the centre of the conductor is zero (since r = 0) and it increases linearly with the increase in distance r within the conductor. The maximum value of field is at the surface of the conductor (corresponding to r = R) and is equal to μ₀I/2πR.

The direction of the field lines is clockwise as we have considered the current to flow into the plane of the figure, away from the reader.

The magnitude of magnetic flux density at a point such as P’ outside the conductor (at distance r > R) is given by

2πrB = µ₀I since the entire current I passes (pierces) through the surface enclosed by the circular path of radius r. This gives

B = μ₀I/2πr

[This is the usual expression for the magnetic field due to a long straight (thin) current carrying conductor].

The variation of the magnitude of the magnetic flux density B against the distance r is shown graphically in the adjoining figure.

Now we will consider the case of a long, straight, thick cylindrical current carrying pipe of inner radius R₁ and outer radius R₂.

[Derivation of the magnetic field due to a current carrying pipe using Ampere’s circuital law was one of the free response questions in the AP Physics C 2011 question paper

A cross section of the pipe is shown in the adjoining figure. The current I is supposed to flow into the plane of the figure, away from the reader.

Consider a point such as P at distance r such that R₁ < r < R₂. To find out the magnetic flux density B at the point P within the material of the pipe, we consider the line integral of B over a circular path (red dotted circle) of radius r and apply Ampere’s circuital law to obtain

∫B. dℓ = µ₀I₁ ………….(i)

where I₁ is the current passing (piercing) through the surface enclosed by the circular path of radius r.

We have I₁ = Iπ(r² – R₁²) /π(R₂² – R₁²) = I(r² – R₁²) /(R₂² – R₁²)

[The total current I passes through the total cross section area π(R₂² – R₁²). So the current I₁ passing through the cross section area π(r² – R₁²) is Iπ(r² – R₁²) /π(R₂² – R₁²) = I(r² – R₁²) /(R₂² – R₁²)].

Substituting for I₁ in equation (i) and remembering that the direction of the magnetic field B is along the path of integration everywhere, we have

2πrB = µ₀I(r² – R₁²) /(R₂² – R₁²). This gives

B = µ₀I(r² – R₁²) /2πr(R₂² – R₁²)

Or, B = (µ₀I/2πr) [(r² – R₁²)/(R₂² – R₁²)]

This expression shows that the magnetic field is zero at r = R₁ (at the inner surface of the pipe) and it increases non-linearly (unlike in the case of a solid cylindrical conductor), within the material of the pipe.

Now, what is the magnetic field inside the hollow region of the pipe? You might have definitely come across this question earlier and might have proved (by other methods) that it is zero everywhere within the hollow region. Well, Ampere’s circuital law offers an immediate proof to you:

When you consider the line integral over a concentric circular path (of the type we considered above) within the hollow region, you find that there is no current passing (piercing) through the surface enclosed by the circular path. Therefore we have

∫B. dℓ = µ₀×0 so that B = 0

What is the magnetic field at a point outside the pipe?

We consider a point such as P’ at distance r > R₂. The magnitude of magnetic flux density B at the point P’ is given by

2πrB = µ₀I since the entire current I passes (pierces) through the surface enclosed by the circular path of radius r. This gives

B = μ₀I/2πr

[As you might have expected, this is same as the usual expression for the magnetic field due to a long straight (thin) current carrying conductor].

The nature of variation of the magnitude of the magnetic field B with the distance r is shown in the adjoining figure.

You will find an interesting question with solution in this section here.