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 B & C - Multiple Choice Practice Questions on Simple Harmonic Motion

Equations to be Remembered in respect of oscillations and simple harmonic motion were discussed in the post dated 17^th April 2008. This was followed by some multiple choice practice questions (with solution) and a free response practice question in the posts dated 22^nd April 2008 and 2^nd May 2008 respectively. A few multiple choice practice questions (with solution) on simple pendulum also were given later in the post dated 12^th September 2009. You can access all those posts by clicking on the label ‘oscillation’ or ‘simple harmonic motion’ below this post.

Today we will discuss a few more multiple choice practice questions (with solution) on simple harmonic motion. The following questions are meant for AP Physics B as well as AP Physics C aspirants:

(1) A girl is swinging on a swing in the sitting position. If she stands up, the the period of the swing will

(1) remain unchanged

(2) increase

(3) decrease

(4) become unpredictable

When the girl stands up, her centre of gravity is elevated and the effective length of the pendulum is decreased. The period of oscillation is therefore decreased. [Option (3)].

(2) The maximum velocity of a particle executing simple harmonic motion with an amplitude 6 cm, is 3.14 ms^–1. The period of oscillation is

(1) 120 s

(2) 12 s

(3) 1.2 s

(4) 0.12 s

The maximum velocity v_max is given by

v_max = Aω where A is the amplitude and ω is the angular frequency.

Therefore, ω = v_max /A

Or, 2π/T = v_max /A from which T = 2πA/ v_max

Substituting known values, period T = 2π×6×10^–2/3.14 = 0.12 s

(3) One and of a light spiral spring is attached to a hook on the ceiling. A mass m kg hung on the spring stretches it by 10 cm. The mass is pulled down a little and released. The period oscillation of the system in seconds is (take g = 10 ms^–2)

(1) 2πm/5

(2) πm/5

(3) π/5

(4) 5π

The spring constant k is given by

k = mg/x = (m×10)/(0.1) = 100m

[Note that the extension x given in centimetre is converted into metre].

The period of oscillation (T) is given by

T =2π√(m/k) = 2π√(m/100m) = 2π/10 = π/5.

(4) You know that the period of oscillation T of a mass m attached to a light spring of force constant k is given by

T =2π√(m/k)

The period of oscillation of such a spring-mass system is found to be 2 s. If the period becomes 3 s when the mass is increased by 2 kg, what is the value of m?

(1) 0.8 kg

(2) 1 kg

(3) 1.2 kg

(4) 1.6 kg

Before adding the extra mass, we have

2 = 2π√(m/k)………….(i)

After adding the extra mass of 2 kg, we have

3= 2π√[(m+2)/k]………(ii)

Dividing Eq (i) by Eq (ii) we have

2/3 = √[m/(m+2)]

Squaring, 4/9 = m/(m+2).

This gives m = 1.6 kg.

The following question is specifically for AP Physics C aspirants:

(5) A (1) A particle executing simple harmonic motion along the y-axis has zero displacement at time t = 0. The period of the motion is 1 s. After what time will its kinetic energy be 25% of the total energy?

(1) 1/12 s

(2) 1/6 s

(3) ¼ s

(4) 1 s

Since the displacement y is zero initially, the equation of the motion is

y = A sin ωt where A is the amplitude and ω is the angular frequency.

The velocity at the instant t is given by

v = dy/dt = Aω cosωt

The kinetic energy at the instant t is given by

½ mv² = ½ mA²ω²cos²ωt where m is the mass of the particle.

The maximum kinetic energy which is equal to the total energy of the particle is ½ mA²ω².

As the kinetic energy at time t is to be equal to 25% of the total energy, we have

½ mA²ω²cos²ωt = ¼ ×½ mA²ω²

This gives cosωt = ½ from which ωt = π/3.

Or, 2πt/T = π/3

Since T = 1 s, we get t = 1/6 s [Option (2)].

You will find additional multiple choice questions in this section here.

AP Physics B & C - Multiple Choice Practice Questions on Simple Pendulum

Equations to be remembered in the section ‘oscillations’ were discussed in the post dated 17^th April 2008. Some multiple choice practice questions in this section were discussed in the post dated 22^nd April 2008, followed by a free response practice question in the post dated 2^nd May 2008. You can access all these posts by clicking on the label ‘oscillation’ below this post. Today we will discuss a few multiple choice practice questions involving simple pendulum:

(1) A simple pendulum arranged inside a train has period of oscillation T when the train is at rest. When the train moves along straight horizontal rails with uniform acceleration of x ms^–2, the period of the pendulum is (assuming g = 10 ms^–2)

(a) T ×g^1/2/(g+x)^1/2

(b) T ×g^1/4/(g+x)^1/4

(c) T ×g^1/4/(g²+x²)^1/4

(d) T ×g^1/2/(g²+x²)^1/2

(e) T ×g^1/2/(g²+x²)^1/4

When the train is at rest the period of oscillation T of the pendulum is given by

T = 2π (ℓ/g)^1/2 ………….(i)

where ℓ is the length of the pendulum and g is the acceleration due to gravity.

When the train moves forward with acceleration x ms^–2, an inertial backward force acts on the bob of the pendulum and supplies a backward acceleration of x ms^–2 (fig.). The resultant acceleration of the bob is (g²+x²)^1/2. The period T₁ of the pendulum is now given by

T₁ =2π[ℓ/(g²+x²)^1/2]^1/2……..(ii)

Dividing eqn (ii) by eqn (i) we have

T₁/T = g^1/2/(g²+x²)^1/4 from which
T₁ = T ×g^1/2/(g²+x²)^1/4
The correct option is (e)

[Suppose the train is moving with uniform velocity of x ms^–1. What will be the period? No doubt, T itself since you cannot distinguish between state of rest and uniform motion].

(2) Two simple pendulums A and B have periods 2.1 s and 2 s respectively. They start oscillating at the same time in phase. They will be in phase instantly at the end of

(a) 42 s

(b) 40 s

(c) 22 s

(d) 21 s

(e) 20 s

Suppose the pendulums are in phase instantly at the end of n oscillations of pendulum A. Pendulum B should then execute n+1 oscillations. Therefore we have

n×2.1 = (n+1)×2

Therefore 0.1n = 2 so that n = 20

The time elapsed is therefore 20×2.1 = 42 s [Option (a)].

[The difference between the periods of the two pendulums is 0.1 s. So you require 20 oscillations of pendulum A to obtain a time difference equal to one period (2 s) of pendulum B so that the two pendulums will be instantly in phase].

(3) The bob of a simple pendulum is a hollow metal sphere filled with water. There is a small hole at the bottom of this hollow sphere and water drains out through the hole as the pendulum oscillates. After completely filling the bob with water, if this pendulum is made to oscillate for a long time, its period of oscillation will

(a) increase first and will reach a final constant value

(b) decrease first, reach a minimum value, then increase and will finally settle at the initial value

(c) increase first, reach a maximum value, then decrease and will finally settle at the initial value

(d) decrease first and will reach a final constant value

(e) remain unchanged

The correct option is (c). Initially the centre of gravity of the spherical bob is at its centre since it is completely filled with water. When the water flows out, the centre of gravity of the bob moves gradually downwards, reaches a minimum level and then moves up. When the water is fully drained out, the centre of gravity of the bob once again reaches the centre of the bob and remains there.

The length of the pendulum is the distance between the centre of gravity of the bob and the point of suspension. Therefore, the length of the pendulum gets increased initially, becomes a maximum, then gets decreased and finally settles at the initial value. Therefore, the period of the pendulum will increase first, reach a maximum value, then decrease and will finally settle at the initial value.

(4) A simple pendulum is taken to a location where the acceleration due to gravity is decreased by 0.1 %. If the period of oscillation is to be unaltered

(a) the mass of the bob is to be increased by 0.1 %

(b) the mass of the bob is to be decreased by 0.1 %

(c) the length of the pendulum is to be decreased by 0.316 %

(d) the length of the pendulum is to be decreased by 0.2 %

(e) the length of the pendulum is to be decreased by 0.1 %

Since the period of oscillation T is given by T = 2π√(ℓ/g), the ratio ℓ/g should be unaltered. Therefore, the length ℓ should be decreased by 0.1 %

The following questions are for AP Physics C aspirants only:

(5) The period of a simple pendulum is decreased by 0.02 s when the length of the pendulum is decreased by 1 cm. The original length of the pendulum is nearly (g = 10 ms^–2)

-->

(a) 0.25 m

(b) 0.5 m

(c) 1 m

(d) 1.01 m

(e) 2 m

The period of oscillation T is given by T = 2π√(ℓ/g).

Therefore, ∆T/T = ½ ∆ℓ/ℓ – ½ ∆g/g

[Here ∆T, ∆ℓ, and ∆g represent the increments in the period, length and acceleration due to gravity respectively. We have written the above equation by taking the logarithm of the expression for period and then differentiating it]

The increment in T is – 0.02 s and the increment in ℓ is – 0.01 m (negative signs are because T and ℓ are decreased). There is no change in g.

Therefore we have – 0.02/T = – ½ × 0.01/ℓ

This gives T = 4ℓ.

But T = 2π√(ℓ/g) so that 2π√(ℓ/10) = 4ℓ

Or, π²/10 = 4ℓ

Since π² is nearly equal to 10, we obtain ℓ = 0.25 m, nearly.

AP Physics B & C –Answer to Free-Response Question (for practice) on Simple Pendulum

In the post dated 30^th April 2008, a free-response question on simple pendulum (for practice) was given to you. As promised, I give below a model answer along with the question:

A simple pendulum of length ℓ is set up using a bob of mass m and a string of negligible mass. The string can withstand a maximum tension of 3mg where g is the acceleration due to gravity. The bob is pulled aside so that the string makes a small angle θ₁ with the vertical. On releasing the bob, the pendulum oscillates with period T. Now, answer the following questions:

(a) Calculate the period of oscillation when the length ℓ = 1 m, assuming that the acceleration due to gravity at the place is 10 ms^–2.

(b) If the bob of the pendulum is now immersed in a liquid of negligible viscosity and of density lower than that of the bob, how will the period be affected? Put a tick mark against the correct option:

Increased ___ Decreased ___ Unchanged ___

Give reason for your answer.

(c) The pendulum bob is taken out from liquid and is oscillated in air itself. Explain what modification you will make to halve the period of oscillation.

(d) The centre of gravity of the bob is raised through a height of 6 cm from its mean position (lowest position) when the bob is in the extreme position during its oscillation. Calculate the speed of the bob when it crosses the mean position.

(e) Calculate the maximum angle through which the string can be displaced from the vertical so that the string will not break when the pendulum oscillates.

(a) The period of oscillation of a simple pendulum of length ℓ is given by

T = 2π√( ℓ/g) where g is the acceleration due to gravity.

Substituting for ℓ (=1 m) and g (= 10 ms^–2), T = 2π√(0.1) = 1.9869 s.

(b) Within the liquid the bob experiences the force of buoyancy exerted by the liquid so that the effective weight of the bob is decreased. The restoring force on the bob is therefore reduced and the period of oscillation is increased.

(c) Since the period is directly proportional to the square root of the length of the pendulum, the length is to be reduced to a quarter of the original length for making the period half the original value.

(d) In the extreme position of the bob, its energy is entirely gravitational potential energy which is mgh where h is the height through which the bob is raised from its mean position. In the mean position this potential energy is completely converted into kinetic energy so that we have

mgh = ½ mv² where v is the speed of the bob at the mean position.

Thus v = √(2gh) = √(2×10×0.06) = 1.0954 ms^–1.

(e) The forces acting on the bob in any position are the weight of the bob mg (which acts vertically downwards) and the tension in the string (which acts radially towards the centre of the arc of the circle along which the bob moves).

The resultant force acting radially towards the centre (point of suspension) supplies the centripetal force for the circular motion so that we have

T – mg cosθ = mv²/ℓ

[Note that the radius of the arc of the circle is the length ℓ of the pendulum].

Therefore, T = mg cosθ + mv²/ℓ

In the mean position we have θ = 0 so that the above equation becomes

T = mg + mv²/ℓ

Since the maximum tension that the string can withstand is 3mg, we have

3mg = mg + mv²/ℓ, from which mv²/ℓ = 2mg.

Therefore, v² = 2gℓ.

This means that in the extreme position the bob is raised through a height ℓ with respect to its mean position [Since v = √(2gh) with usual notations].

The angle which the string makes with the vertical is therefore 90º in the limiting case. [In other words this is the maximum angle through which the string can be displaced from the vertical so that it will not break when the pendulum oscillates].

* * * * * * * * * * * * * * * * * * * *

[Suppose the string would break when the tension is 2mg.

You will then have 2mg = mg + mv²/ℓ, from which mv²/ℓ = mg.

Therefore, v² = gℓ. Since v² = 2gh , h = ℓ/2.

But h = ℓ(1– cosθ₁) where θ₁ is the maximum angle through which the string can be displaced from the vertical. Therefore, (1– cosθ₁) = ½ from which θ₁ = 60º].