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

AP Physics B & C – Multiple Choice Questions (for practice) on Direct Current Circuits

The essential formulae to be remembered in connection with direct current circuits were given in the post daied 31^st March 2008. A couple of free response questions were posted on 5^th April 2008. Today we will discuss a few multiple choice questions in this section.

In the circuit shown, the battery has some internal resistance. Answer the following questions numbered (1) and (2) in respect of the circuit:

Question No.(1):

The switch S has been closed for a long time and the capacitor C has been fully charged. If the voltage across the capacitor is found to be 3 V, what is the current flowing through the resistor R₁?

(a) zero

(b) 0.2 A

(c) 0.5 A

(d) 0.6 A

(e) 6/11 A

Since the capacitor is fully charged, the voltage across the capacitor is equal to the steady voltage across the 6 Ω resistor R₃. The current I flowing through R₃ is therefore given by

I = 3 V/6 Ω = 0.5 A.

The current through R₁ is the same as that through R₃. So the correct option is 0.5 A.

Question No.(2):

What is the internal resistance of the battery?

(a) 0.2 Ω

(b) 0.5 Ω

(c) 1 Ω

(d) 18 Ω

(e) 6/11 A

The current I delivered by the battery can be written as

I = E/(R₁+ R₂+ R₃+r) where E is the emf and r is the internal resistance of the battery. Since the current I is 0.5 ampere, we have

0.5 A = 9V/(5+6+6+r)Ω so that r = 1 Ω.

Question numbers (3), (4) and (5) given below relate to the adjoining circuit in which a current of 2A enters the network of resistors R₁, R₂, R₃, R₄ and R₅ through the junction P and leaves through the junction Q.

Question No.(3):

The effective resistance between the junctions P and Q is

(a) 1 Ω

(b) 2 Ω

(c) 3 Ω

(d) 4 Ω

(e) 5 A

The resistors R₁, R₂, R₃ and R₄ make a balanced Wheatstone bridge since R₁/R₂ = R₃/R₄. No current can therefore pass through the resistor R₅ so that it can be ignored. The circuit therefore reduces to 3Ω (which is the series combined value of 1Ω and 2Ω) in parallel with 6Ω (which is the series combined value of 2Ω and 4Ω). Therefore, the effective resistance between the junctions P and Q is (3×6)/(3+6) = 2 Ω.

Question No.(4):

The potential difference between the junctions P and Q is

(a) 2 V

(b) 4 V

(c) 6 V

(d) 8 V

(e) 10 V

Since the effective resistance between the junctions P and Q is 2 Ω (as shown above) and the current flowing into the network is 2 A, the potential difference (IR drop) between the junctions P and Q is 2×2 = 4 V.

Question No.(5):

The current flowing through the 4 Ω resistor (R₄) is very nearly

(a) 0.567A

(b) 0.667 A

(c) 0.876 A

(d) 0.95 A

(e) 1 A

The resistor R₅ can be ignored as we saw while answering question No.3. The current flowing through R₄ is the current flowing through the lower parallel branch containing R₃ and R₄. The lower parallel branch has total resistance 6 Ω and the upper branch has total resistance 3 Ω. The current being inversely proportional to the resistance of the branch, the main current of 2A must get divided in the ratio 3:6 (= 1:2) between the lower and upper branches. The current through the lower branch must therefore be 2×(1/3) =2/3 A which is very nearly equal to 0.667 A.

[When current gets divided between two parallel branches, the current (I₁) through one branch is given by

I₁ = (Main current× Resistance of the other branch)/ Total resistance

In the present case current through the branch containing R₄ = 2× (R₁+R₂)/ (R₁+R₂+ R₃+R₄) =

2× (1+2)/(1+2+2+4) = 2/3 = 0.667 A, very nearly].

You will find similar multiple choice questions (MCQ) with solution from different branches of physics at physicsplus.blogspot.com

AP Physics B & C –Answers to Free-Response Questions (for practice) on Direct Current Circuits

Two free response questions on direct current circuits were posted on 5^th March 2008 for your practice. As promised, I give below the answers along with the questions.

Free Response Question No.1 (For AP Physics B and C)

A 10 V battery of negligible internal resistance is connected in series with a 4 Ω resistor R₁, a 6 Ω resistor R₂ and a 2 µF capacitor C as shown in the figure. A series combination of a 6 Ω resistor R₃ and a switch S is connected across the series combination of R₂ and C. The switch S is initially open and the battery has been connected for a long time.

(a) Determine the voltage across the capacitor C and the magnitude of the charge on one of its plates.

(b) The switch S is now closed. Determine the voltage across the capacitor C immediately after closing the switch S.

(c) Calculate the voltage across the capacitor C a long time after closing the switch S.

(d) Calculate the energy stored in the capacitor C a long time after closing the switch S.

(e) On closing the switch S, will the energy stored in the capacitor increase, decrease or remain unchanged? Put a tick mark against the correct option

Increase _____ Decrease _____ Remain unchanged _____

Justify your answer.

(a) Voltage across the capacitor is the same as the battery emf 10 V since the capacitor will charge through the resistors R₁ and R₂ to the full applied voltage on connecting it to the source for a long time.

(b) Immediately after closing the switch, the voltage across the capacitor is 10 volts it self since the capacitor cannot discharge immediately through R₂ and R₃.

(c) When the switch S is closed, the capacitor starts discharging through R₂ and R₃. After a long time, the voltage across the capacitor becomes equal to the voltage across R₃.

Voltage across R₃ = (Current through R₃)×R₃ = [10/(6+4)]×6 = 6 V.

(d) The energy stored in the capacitor = ½ CV² = ½ ×(2×10^–6)×6² = 36×10^–6 coulomb = 36 μC.

(e) The energy will decrease since the voltage across the capacitor decreases from 10 V to 6 V.

Free Response Question No.2 (For AP Physics C)

In the adjoining figure the capacitor C is initially uncharged and the switches S₁ and S₂ are open. The battery has negligible internal resistance.

(a) The switches S₁ and S₂ are closed simultaneously. What will be the voltage across the resistor R₂ immediately after closing the switches S₁ and S₂? Justify your answer.

(b) Determine the current through the resistor R₂ immediately after closing the switches S₁ and S₂.

(c) A long time after closing the switches S₁ and S₂, the voltage across the resistor R₁ was found to be 6 V. Calculate the value of the resistor R₂.

(d) After a long time the switch S₁ is opened but the switch S₂ is retained in the closed condition. Calculate the voltage across the resistor R₂ at the instant when 0.6 seconds have been elapsed after opening switch S₁.

(a) The capacitor draws the maximum charging current initially and therefore effectively places a short circuit across the resistor R₂ and the entire supply voltage of 9 volts appears across R₁.

[This is the same as saying that the initial potential difference of zero volt across the capacitor cannot change abruptly since it takes time for the charges to get accumulated on the plates of the capacitor and to increase the P.D. across the plates].

(b) Since the initial voltage across the resistor R₁ is 9 volts, the initial current through it (immediately after closing the switches S₁ and S₂) is 9 V/12 KΩ = 0.75 mA.

(c) The current flowing through R₁ = 6V/12 KΩ = 0,5 mA.

The voltage across R₂ = 9 V – 6 V = 3 V.

Since the same current of 0.5 mA flows through R₁ and R₂ we have

R₂ = 3 V/0.5 mA = 6 KΩ.

(d) During the discharge of a capacitor through a resistance R, the charge Q on the capacitor at time t is given by Q =Q₀ e^–t/RC where Q₀ is the initial charge. Since V = Q/C, the voltage V across the capacitor is given by

V = (Q₀/C) e^–t/RC = V₀ e^–t/RC where V₀ is the initial voltage which is 3 volts.

After 0.6 seconds, the voltage across the capacitor (which is also the voltage across R₂) is given by

V = 3×e^–0.6/0.6, since RC = R₂C = 6000 Ω × 100×10^–6 F = 0.6 second.

Thus V = 3×e^–1 = 1.104 volt (very nearly)

AP Physics B & C –Two Free-Response Questions (for practice) on Direct Current Circuits

Free response questions help the examiner to assess your standard from different angles and to some extent the evaluation will be subjective. But you can rest assured that the questions are so set that they are as specific as practicable. You must be very careful in presenting your answers in respect of free-response questions. Since the words “determine”, “derive”, “calculate”, “ justify”, “explain”, “what is”, “sketch” and “plot” have precise meanings in the AP Physics examination, you must thoroughly understand their significance so that you can give enough details wherever required and refrain from giving unwanted details where they will be sheer waste of time.

Here are two typical free response questions on direct current circuits. The first question is meant for both AP Physics B and C while the second question is meant for AP Physics C only.

(1) A 10 V battery of negligible internal resistance is connected in series with a 4 Ω resistor R₁, a 6 Ω resistor R₂ and a 2 µF capacitor C as shown in the figure. A series combination of a 6 Ω resistor R₃ and a switch S is connected across the series combination of R₂ and C. The switch S is initially open and the battery has been connected for a long time.

(a) Determine the voltage across the capacitor C and the magnitude of the charge on one of its plates.

(b) The switch S is now closed. Determine the voltage across the capacitor C immediately after closing the switch S.

(c) Calculate the voltage across the capacitor C a long time after closing the switch S.

(d) Calculate the energy stored in the capacitor C a long time after closing the switch S.

(e) On closing the switch S, will the energy stored in the capacitor increase, decrease or remain unchanged? Put a tick mark against the correct option

Increase _____ Decrease _____ Remain unchanged _____

Justify your answer.

The above question carries 15 points.

(2) In the adjoining figure the capacitor C is initially uncharged and the switches S₁ and S₂ are open. The battery has negligible internal resistance.

(a) The switches S₁ and S₂ are closed simultaneously. What will be the voltage across the resistor R₂ immediately after closing the switches S₁ and S₂? Justify your answer.

(b) Determine the current through the resistor R₂ immediately after closing the switches S₁ and S₂.

(c) A long time after closing the switches S₁ and S₂, the voltage across the resistor R₁ was found to be 6 V. Calculate the value of the resistor R₂.

(d) After a long time the switch S₁ is opened but the switch S₂ is retained in the closed condition. Calculate the voltage across the resistor R₂ at the instant when 0.6 seconds have been elapsed after opening switch S₁.

This question carries 15 points.

Try to answer the above questions. I’ll be back shortly with model answers for you.