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].

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

The following free response question will be useful for both AP Physics B and C aspirants:

A simple pendulum of length ℓ is set up using of 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.

Try to answer the above question which carries 15 points which can be distributed among the parts (a), (b), (c), (d) and (e) as 2+3+2+4+4. You can take about 17 minutes for answering the above question.

I’ll be back with a model answer for your benefit shortly.

AP Physics B & C –Multiple Choice Questions on Oscillations and Simple Harmonic Motion

Let us discuss some typical multiple choice questions on oscillations and simple harmonic motion.

(1) A spring of negligible mass and force constant k is suspended from a rigid support and a mass m is attached to its free end. On depressing the mass slightly and releasing, the system executes simple harmonic oscillations of frequency f. The spring is now cut into two pieces with lengths in the ratio 1:2. If the mass m is attached to the longer piece, the frequency of oscillations will be

(a) 2f

(b) (2/3) f

(c) (3/2) f

(d) √(3/2) f

(e) √(2/3) f

The period of oscillation of a spring mass system is given by

T = 2π√(m/k)

where k is the force constant of the spring. The frequency of oscillations is therefore given by

f = (1/2π) √(k/m)

When the spring is cut in the ratio (of lengths) 1:2, the force constants of the pieces are 3k and (3/2)k.

[You should note that the spring constant is inversely proportional to the length of the spring. Therefore, when the length of the spring becomes one-third, the spring constant becomes three times (3k) and when the length becomes two-thirds, the spring constant becomes (3/2)k].

The frequency therefore becomes (1/2π) √(3k /2m) = √(3/2) f.

(2) One end of a spring of force constant k is fixed to a rigid support S on an inclined plane of angle θ and the other end is connected to a block of mass m as shown in the adjacent figure. The inclined plane is smooth. If the block is displaced along the plane through a small distance and released, the mass executes simple harmonic oscillations. The period of oscillations is

(a) 2π√(m/k)

(b) 2π√(m/k sinθ)

(c) 2π√(m/k cosθ)

(d) 2π√(m sinθ/k)

(e) 2π√(mg sinθ /k)

The period of oscillation of the spring mass system is independent of the gravitational field since the restoring force is supplied by the elastic force in the spring. The period is therefore independent of the direction of motion and is given by

T = 2π√(m/k)

(3) A wooden cube of side a and mass m floats in a liquid of density ρ. On giving a small vertical displacement from the equilibrium position, the cube oscillates with a period T. Then T is

(a) directly proportional to aρ

(b) directly proportional to √(aρ)

(c) inversely proportional to √(aρ)

(d) inversely proportional to a√ρ

(e) independent of a

The period of oscillation is given by

T = 2π√(m/k) where m is the mass of the cube and k is the force per unit displacement of the block.

When you push the floating cube down through a small distance x, the additional force of buoyancy on the cube is equal in magnitude to the weight of the additional liquid displaced by the cube. This is vρg where v = extra volume of liquid displaced = a²x.

Thus, the restoring force on the cube = vρg = a²xρg.

Therefore, the force per unit displacement of the cube, k = a²xρg/x = a²ρg

The period of oscillation is thus given by T = 2π√(m/k) = 2π√(m/a²ρg)

The period is therefore inversely proportional to a√ρ [Option (d)]

[Note that the period of oscillation is 2π√(m/Aρg) where A is the area of cross section. This expression holds in the case of any block of uniform cross section area A].

(4) A simple pendulum is arranged inside an elevator. When the elevator moves upwards with a uniform velocity of 1 ms^–1, the period of oscillation is T. If the elevator moves upwards with an acceleration of 1 ms^–2, the period of oscillation of the pendulum will be (assuming g = 10 ms^–2)

(a) T ×√(10/11)

(b) T ×√(11/10)

(c) T ×√11)

(d) T ×√10

(e) T

The period of oscillation of simple pendulum is given by

T = 2π√(ℓ/g) where ℓ is the length of the pendulum and g is the acceleration due to gravity. The period of the pendulum is given by the above equation when the elevator is at rest and also when it is moving with uniform velocity.

Since g = 10 ms^–2, we have T = 2π√(ℓ/10).

When the elevator moves up with acceleration (a) of 1 ms^–2, the weight of the bob of mass m increases from mg to m(g+a) so that the restoring force on the bob increases correspondingly. The period of oscillation therefore decreases.

In place of g the value (g+a) is to be substituted in the expression for period.

The modified period (T₁) is given by

T₁ = 2π√[ℓ/(10+1)] = 2π√(ℓ/11) = 2π√[(ℓ/10) ×(10/11)] = T ×√(10/11).

(5) The displacement (y) of a particle is given by the equation,

y = 5 (cos² 3πt – sin² 3πt). The motion of the particle is

(a) not simple harmonic

(b) simple harmonic with frequency 1 Hz

(c) simple harmonic with frequency 3 Hz

(d) simple harmonic with frequency 6 Hz

(e) simple harmonic with frequency 3π Hz

If you are a little bit strong in trigonometry, you will get the correct option (c) easily. We have cos² A– sin² A = cos 2A so that the equation for the displacement of the particle is

y = 5 cos 6πt, which is in the form y = A cos ωt

Therefore, the angular frequency ω = 6π. The linear frequency f = ω/2π = 3 Hz.

Practice makes one perfect. So try to work out as many questions as possible to make you confident to face the AP Physics Exam. You will find some useful multiple choice questions in this section at physicsplus: Additional Multiple Choice Questions on Simple Harmonic Motion

AP Physics B & C – Oscillations and Simple Harmonic Motion – Equations to be Remembered

You must remember the following points to make you strong in answering multiple choice questions involving oscillations and simple harmonic motion :

(1) The simplest equation of simple harmonic motion is

y = Asinωt if initial phase and displacement are zero. Here ‘y’ is the displacement, ‘ω’ is the angular frequency and A is the amplitude.

y = Acosωt also represents simple harmonic motion but it has a phase lead of π/2 compared to the above one.

If there is an initial phase of Φ the equation is

y = Asin(ωt + Φ).

[Or, y = Acos(ωt + Φ), if you use the cosine form]

y = Asinωt + Bcosωt represents the general simple harmonic motion of amplitude √(A² + B²) and initial phase tan^-1(B/A).

The above equation can be modified by putting

A = A₀cos Φ and

B = A₀sin Φ to yield

y = A₀cos Φ sinωt + A₀sin Φ cosωt = A₀sin(ωt + Φ) which is the standard equation of a simple harmonic motion. [Evidently, A²+B² = A₀² and B/A = tan Φ].

(2) The differential equation of simple harmonic motion is

d²y/dt² = -ω²y

Note that ω =√(k/m) where ‘k’ is the force constant (force per unit displacement) and ‘m’ is the mass of the particle executing the SHM.

(3) Velocity of the particle in SHM, v = ω√(A² – y²)

Maximum velocity, v_max = ωA

(4) Acceleration of the particle in SHM, a = - ω²y

Maximum acceleration, a_max = ω²A

(5) Kinetic Energy of the particle in SHM, K.E. = ½ m ω²( A² –y²)

Maximum Kinetic energy = ½ m ω²A²

Potential Energy of the particle in SHM, P.E. = ½ m ω²y²

Maximum Potential Energy = ½ m ω²A²

Total Energy in any position = ½ m ω²A²

Note that the kinetic energy is maximum in the mean position and the potential energy is maximum in the extreme position. The sum of the kinetic and potential energies which is the total energy is a constant in all positions. Remember this:

Maximum K.E. = Maximum P.E. = Total Energy = ½ m ω²A²

(6) Period of SHM = 2π√(Inertia factor/ Spring factor)

In cases of linear motion as in the case of a spring-mass system or a simple pendulum, period, T = 2π√(m/k) where ‘m’ is the mass and ‘k’ is the force per unit displacement.

In the case of angular motion, as in the case of a torsion pendulum,

T = 2π√(I/c) where I is the moment of inertia and ‘c’ is the torque (couple) per unit angular displacement.

You may encounter questions requiring calculation of the period of seemingly difficult simple harmonic oscillators. Understand that the question will become simple once you are able to find out the force constant in linear motion and torque constant in angular motion. You will usually encounter cases of linear simple harmonic motion and it won’t be difficult to find he force constant and the period.

(7) In the case of the oscillations of a mass m on a spring of negligible mass, the inertia factor is the mass m attached to the spring and the spring factor is the force constant (spring constant) k of the spring so that the period of oscillation is given by

T = 2π√(m/k)

If two springs of spring constants k₁ and k₂ are connected in series as shown, the effective spring constant k is given by the reciprocal relation,

1/k = 1/k₁ + 1/k₂ so that k = k₁k₂/(k₁+k₂)

[If many springs are connected in series, you will write 1/k = 1/k₁ + 1/k₂ +1/k₃ + ……etc.]

If springs are connected in parallel as shown in the figure, the effective spring constant will be the sum of the individual spring constants.

If springs are connected on opposite sides of a mass as shown, again the effective spring constant is the sum of the individual spring constants.

If two masses (m₁ and m₂) are connected by a spring of force constant k and the system is placed on a smooth surface, on compressing the spring by pushing the masses towards each other simultaneously and releasing, the masses oscillate with a period

T = 2π√(m/k) where the effective mass m = m₁ m₂ /(m₁ + m₂)

(8) 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.

[Note that the period of oscillation of a spring mass system is independent of the acceleration due to gravity, unlike the simple pendulum].

Questions in this section will be discussed in the next post.

Meanwhile, find many useful multiple choice questions (MCQ) with solution from different branches of physics at Physicsplus