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.

Answer to Free Response Practice Question on Kinematics in Two Dimensions for AP Physics C

A free response practice question on kinematics in two dimensions was given to you in the post dated 11^th February 2010. As promised, I give below a model answer along with the question:

A chopper at an air base is rising vertically up with a velocity of 2 ms^–1. When it is at an altitude of 50 m, a packet is thrown down with velocity 8 ms^–1 (with respect to the chopper), making an angle of 60º with the vertical. Neglecting air resistance and assuming that g = 10 ms^–2, answer the following questions

(a) Draw a diagram to show the nature of the path AB followed by the packet from the moment it leaves the chopper to the moment it hits the ground, as seen by an observer on the ground. Is the path straight, circular or parabolic?.

(b) Will the packet move up initially? Justify your answer.

(c) Calculate the time taken by the packet to reach the ground.

(d) Determine the horizontal distance traveled by the packet before hitting the ground.

(e) Determine the velocity (magnitude as well as direction) with which the packet hits the ground.

(a) The path AB of the packet is shown in the following figure. The path is parabolic.

(b) The packet will not move up initially. This is due to the fact that the vertical, downward component (8 cos 60º = 4 ms^–1) of the velocity with which the packet is thrown down is greater than the upward velocity (2 ms^–1) of the chopper.

[The situation is as though the packet is thrown down from a stationary chopper with vertical (downward) velocity component (4 ms^–1 – 2 ms^–1) = 2 ms^–1 and horizontal velocity component 8 sin 60º = 4√3 ms^–1].

(c) The time t taken by the packet to reach the ground can be found by considering its vertical motion. We have

s = ut + ½ at² where the vertical displacement s = 50 m, vertical (downward) velocity component u = 8 cos 60º – 2 = 2 ms^–1 and vertical (downward) acceleration a = g = 10 ms^–2.

Therefore, 50 = 2t + ½ ×10t².

Or, 5t² + 2t – 50 = 0

This gives t = [–2 ± √(4 – 1000)] /10 = 2.968 s.

(d) The horizontal distance R traveled by the packet before hitting the ground is the product of the above time and the horizontal velocity component v_horizontal which is equal to 8 sin 60º (= 4√3 ms^–1).

Therefore, R = 2.968×4√3 = 20.56 m.

(e) The velocity with which the packet hits the ground is the resultant of the horizontal and vertical components. The horizontal component of velocity remains unchanged (at 8sin 60º = 4√3 ms^–1) throughout the motion of the packet since the gravitational force acts vertically and cannot affect the horizontal motion. The vertical component of velocity v_vertical goes on increasing during the fall of the packet in accordance with the equation,

v_vertical = u_vertical + at where u_vertical is the initial vertical component of velocity, a is the vertical acceleration and t is the time of flight of the packet.

Here u_vertical = 8 cos 60º – 2 = 2 ms^–1, a = g = 10 ms^–2 and t = 2.968 s.

Therefore, v_vertical = 2 + 10×2.968 = 31.68 ms^–1.

The magnitude v of the velocity with which the packet hits the ground is therefore given by

v = √[(4√3)² + (31.68)²] = 32.43 ms^–1.

The direction of this velocity makes an angle θ with the horizontal (fig.) and is given by

tan θ = v_vertical/ v_horizontal = 31.68/(4√3) = 4.572

Therefore θ = 77.66º

Kinematics in Two Dimensions – Free Response Practice Question for AP Physics C

Today I’ll give you a free response practice question on Kinematics in Two Dimensions. Even though this question is meant for AP Physics C aspirants, those who prepare for AP Physics B exam also will find it useful. Here is the question:

A chopper at an air base is rising vertically up with a velocity of 2 ms^–1. When it is at an altitude of 50 m, a packet is thrown down with velocity 8 ms^–1 (with respect to the chopper), making an angle of 60º with the vertical. Neglecting air resistance and assuming that g = 10 ms^–2, answer the following questions

(a) Draw a diagram to show the nature of the path AB followed by the packet from the moment it leaves the chopper to the moment it hits the ground, as seen by an observer on the ground. Is the path straight, circular or parabolic?.

(b) Will the packet move up initially? Justify your answer.

(c) Calculate the time taken by the packet to reach the ground.

(d) Determine the horizontal distance traveled by the packet before hitting the ground.

(e) Determine the velocity (magnitude as well as direction) with which the packet hits the ground.

Try to answer this question.. You have 15 minutes at your disposal and can score up to 15 points for the right answer. Of course I’ll be back soon with a model answer for your benefit.

Kinematics in One Dimension –Multiple Choice Practice Questions for AP Physics B & C

All posts related to kinematics on this blog can be accessed either by clicking on the label ‘kinematics’ below this post or by performing a search using the search box at the top of this page. Today I give you a few multiple choice practice questions with solution:

(1) A stone projected vertically up is found to be at a height h at times 1 s and 3 s. Neglect air resistance and assume that g = 10 ms^–2. The maximum height reached by the stone is

(a) 60 m

(b) 50 m

(c) 44 m

(d) 22 m

(e) 20 m

Evidently the stone reaches the same height h first during its upward trip and next during its downward trip. But you need not bother about this because everything is contained in the equation of motion, x – x₀ = v₀t + ½ at² (or, s = ut + ½ at²)

Considering the two cases given in the question, since a = – g (taking upwrd as positive and therefore downward negative), we have

h = v₀×1 – ½ ×10×1² = v₀×3 – ½ ×10×3² from which

2v₀ = 40 so that v₀ = 20 ms^–1

The naximum height h_max reached is given by by 0 = v₀² – 2gh_max

[This is the equation, v² = v₀² + 2a(x – x₀) which can also be written as v² = u² + 2as]

Therefore, h_max = v₀²/2g = 20²/(2×10) = 20 m

(2) A food packet is released from a helicopter ascending vertically up with uniform velocity of 2 ms^–1. Neglect air resistance and assume that g = 10 ms^–2. The velocity of the packet at the end of 0.2 s will be

(a) 5 ms^–1

(b) 4 ms^–1

(c) 2 ms^–1

(d) 1 ms^–1

(e) zero

The initial velocity v₀ of the packet is 2 ms^–1 and is directed upwards. The gravitational acceleration is directed downwards. The velocity v of the packet at time t is given by

v = v₀ – gt.

[Note that we have taken the upward vector as positive and hence the downward vector as negative].

Substituting for v₀, g and t, v = 2 – (10×0.2) = 0.

[After getting released, the packet moves upwards for 0.2 s with uniformly decreasing speed, momentarily comes to rest and then moves downwards with uniformly increasing speed].

(3) An astronaut on a planet having no atmosphere throws a stone of mass m vertically upwards with velocity 10 ms^–1. It reaches maximum height of 10 m. If another stone of mass 2m is thrown upwards with velocity 20 ms^–1, the maximum height reached will be

(a) 80 m

(b) 40 m

(c) 30 m

(d) 20 m

(e) 10 m

The maximum height (h_max) reached is given by

h_max = v₀²/2g

[Remember v² = v₀² + 2a(x – x₀) which can be written as v² = v₀² – 2gh_max for projection vertically apwards]

Substituting for h_max (= 10 m) and v₀ (= 10 ms^–1), the acceleration due to gravity (g) on the planet is obtained as 5 ms^–2.

For a given velocity of projection, the height reached is independent of the mass of the stone.

When the velocity of projection is changed to 20 ms^–1, the maximum height reached is 20²/(2×5) = 40 m.

[You may note that the maximum height is directly proportional to the square of the velocity of projection. If the velocity of projection is doubled, the maximum height reached must be quadrupled as in the present problem].

The following question is specifically for AP Physics C aspirants. But AP Physics B aspirants too can ‘enjoy’ it.

(4) A ball thrown vertically up from the ground moves very close to an open window of a room in the first floor of a building. A student in the room sees the ball through the window, when the ball moves up as well as when it move down. The height of the window is 2 m and the ball is in view for a total time of 0.2 s. Take g = 10 ms^–2. How high above the top of the window does the ball go (approximately)?

(a) 19 m

(b) 20 m

(c) 21 m

(d) 24 m

(e) 25 m

Since the total time for which the ball is in view is 0.2 s, the time taken by the ball to traverse the height of the window is 0.1 s. The average upward velocity (v) of the ball while moving past the window is 2/0.1 = 20 ms^–1. We may take this to be very nearly equal to the velocity of the ball at the centre of the window.

Therefore, the maximum height reached by the ball (from the centre of the window) is given by

h_max = v²/2g = 20²/(2×10) = 20 m.

[We have used the equation, v² =u² + 2as where a = – g and s = h_max].

Therefore, the ball goes to a height of nearly 19 m above the top of the window [Option (a)].

[Since the motion of the ball is accelerated, the average velocity is not exactly equal to the velocity at the centre of the window. The answer we obtain is an approximate one.

If you wnt to calculate the exact height, you will proceed as follows:

If t is the time taken by the ball to fall from the topmost point of the trajectory to the top of the window, the maximum height h_max (measured from the top of the window) is given by

h_max = 0 + ½ gt² = 5 t² ………(i)

Also, (h_max + 2) = 0 + ½ g (t +0.1)² = 5t² + 5×0.01 + 5×2×0.1t

Or, h_max + 2 = 5 t² + 0.05 + t ……….(ii)

Subtracting Eq (i) from Eq (ii) we obtain t = 1.95 s.

Therefore, h_max = 5 t² = 5×(1.95)² = 19.0125 m.

AP Physics B- Additional Practice Questions (MCQ) on Geometric Optics

Some multiple choice practice questions on geometric optics were discussed in the posts dated 1^st January 2008 and 3rd January 2008. You can access all posts related to geometric optics by clicking on the label ‘geometric optics’ below this post.

Today we will discuss a few more multiple choice practice questions in this section:

(1) The sun subtends an angle θ radian on the surface of the earth. If a convex lens of focal length F is used to obtain a real image of the sun on a screen, the diameter of the image will be

(a) Fθ

(b) Fθ/2

(c) F/θ

(d) 2Fθ

(e) impossible to be calculated with the given data.

Data is certainly sufficient for calculation. A very simple figure showing the formation of the image of the sun is shown. Since the sun is far away from the lens, its real image is formed at the focus of the lens. The angle subtended by this image at the optic centre of the lens is θ radian so that the diameter d of the image is Fθ.

[The sun subtends an angle of nearly 0.5º on the surface of the earth. If you use a convex lens of focal length 1 m, the diameter of the image will be Fθ = 1×(0.5×π/180), remembering that 180º = π radian. The diameter of the image will be nearly 0.0087 m = 0.87 cm].

(2) A slide projector gives a linear magnification of 20. If it projects a 4 cm×3 cm slide, the area of the image will be

(a) 0.24 m²

(b) 0.48 m²

(c) 4.8 m²

(d) 24 m²

(e) 48 m²

The length as well as the breadth will be magnified 20 times. In other words, the areal magnification is 20×20 = 400. Since the area of the slide is 12 cm², the area of the image will be 12×400 = 4800 cm² = 0.48 m².

(3) Crown glass has refractive index 1.5. A cube of side 12 cm made of crown glass has a small air bubble inside and it appears to be at a distance of 6 cm when viewed normally through one face of the cube. If the air bubble is viewed normally through the opposite face, it will appear to be at a distance of

(a) 6 cm

(b) 5 cm

(c) 4 cm

(d) 3 cm

(e) 2 cm

In the case of normal refraction, the real distance d_real is related to the apparent distance d_app by the equation

n = d_real /d_app where n is the refractive index of the medium in which the object is placed. The refractive index n is with respect to the medium from which the object is viewed.

Therefore we have

1.5 = d_real /6 from which d_real = 9 cm.

Since the real distance of the air bubble from one face is 9 cm, the real distance from the opposite face is (12–9) cm = 3 cm.

Therefore the apparent distance x from the opposite face is given by

1.5 = 3/x from which x = 2 cm [Option (e)].

(4) A thin plano-convex lens has focal length 40 cm. The radius of curvature of its curved surface is 20 cm. If its curved surface is silvered, it can function as a concave mirror of focal length very nearly

(a) 6.67 cm

(b) 8.48 cm

(c) 10 cm

(d) 20 cm

(e) 32.8 cm

When light is incident on the plane face, it is refracted by the plano-convex lens. Then it is reflected by the concave mirror formed by the silvered curved face. While returning, once again it is refracted by the plano-convex lens. In effect the system behaves as the combination of two plano-convex lenses (each of focal length 40 cm) and a concave mirror (of focal length 10 cm since the radius of curvature is 20 cm), obeying the power relation,

1/F = 1/f₁ + 1/f₂ + 1/f₃ where F is the combined focal length and f₁, f₂ and f₃ are the individual focal lengths.

In the present case, f₁ = f₂ = 40 cm and f₃ = 10 cm.

Therefore, 1/F = 1/40 + 1/40 + 1/10 = 1/20 + 1/10 = 3/20

This gives F = 6.67 cm, nearly.

[If the above lens is silvered on its plane face instead of the curved face, the focal length F of the equivalent concave mirror will be given by

1/F = 1/40 + 1/40 + 1/∞ since the focal length of a plane mirror is infinity.

Therefore, F = 20 cm].