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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Tuesday, April 29, 2014

### AP Physics B - Multiple Choice Practice Questions on Wave Motion including Sound

“Being ignorant is not so much a shame, as being unwilling to learn.”
– Benjamin Franklin

Today we shall discuss a few simple multiple choice practice questions on wave motion including sound. Often your knowledge and understanding of basic principles will be tested in the AP Physics Examination and the questions I give below are meant for this.

(1) Here are a few common waves:

(i) Infra red waves (ii) Microwaves (iii) Light waves (iv) Sound waves

Which of the above waves can propagate through vacuum?

(a) (ii) and (iii)

(b) (i) (ii) and (iii)

(c) (1) and (iv)

(d) (iii) and (iv)

(e) None

Infra red waves microwaves and light waves are electromagnetic waves and hence they do not require any medium for their propagation. Sound waves are mechanical waves which require a material medium for their propagation. The correct option is (b).

(2) When a sound source moves past a listener,

(a) the pitch of the sound decreases continuously

(b) the pitch of the sound increases continuously

(c) the pitch of the sound remains unchanged

(d) the pitch of the sound increases suddenly

(e) the pitch of the sound decreases suddenly

The pitch (frequency) of the sound as heard by the listener when the source of sound moves towards the listener, is greater than the actual frequency of the source (in accordance with Doppler effect). The apparent frequency (n1) of the sound in this situation is given by

n1 = nv/(v vS) where n is the actual frequency of the source, v is the speed of sound and vS is the speed of the source.

The pitch (frequency) of the sound as heard by the listener when the source of sound moves away from the listener, is less than the actual frequency of the source. The apparent frequency (n2) of the sound in this situation is given by

n2 = nv/(v+vS)

Therefore, when a sound source moves past a listener, the pitch of the sound decreases suddenly [Option (e)].

[You may click here to see a useful post in which the equations to be noted in this section are given].

(3) A fighter plane moves away from a radar installation at a speed equal to twice the speed of sound. If the real frequency of the sound emitted by the fighter plane is n, what is the apparent frequency of the sound of the plane as heard by an observer at the radar installation?

(a) zero

(b) 3n

(c) n/3

(d)n/2

(e) 2n

This is a case of Doppler effect produced when the source of sound moves away from a listener. The apparent frequency (n’) of the sound in this situation is given by

n’ = nv/(v+vS) ) where n is the actual frequency of the source, v is the speed of sound and vS is the speed of the source.

Since vS = 3v in the present case, we obtain

n’ = n/3, as given in option (c).

(4) Tuning fork A has a small piece of wax attached to one of its prongs (Fig.).  When this fork and another fork B of frequency 286 Hz are excited together, 3 beats per second are produced. The wax on the fork A is now removed and the two forks are again excited together. The number of beats per second is found to be 3 itself. What is the frequency of fork A when the wax on it is removed?
(a) 286 Hz
(b) 289 Hz
(c) 283 Hz
(d) 280 Hz
(e) 292 Hz
The beat frequency is the difference between the frequencies of the forks. Since the fork A without wax produces 3 beasts per second with the fork B of frequency 286 Hz, the frequency of fork A must be either 289 Hz or 283 Hz. If the frequency of A is 283 Hz, its frequency when loaded with wax will be less than 283 Hz and it will produce more than 3 beats per second when excited together with for B. Therefore, the frequency of fork A must be 289 Hz [Option (b)].
[What happens is this:
When the fork A is loaded with wax, its frequency gets reduced from 389 Hz to 383 Hz and it produces 3 beats per second when excited together with fork B of frequency 286 Hz. When the wax on the fork A is removed, its frequency becomes its original frequency 289 Hz and once again it produces 3 beats per second when excited along with fork B of frequency 286 Hz].
(5) A wave has amplitude A given by
A = 2b/(b – c + d)
Then the condition for resonance is
(a) b = d and c = 0
(b) b = 0 and c = d
(c) b = c = d
(d) b = c and d = 0
(e) b = c + d
The amplitude A will be infinite when b = c and d = 0. Therefore the condition for resonance is given in option (d).

## Thursday, April 10, 2014

### AP Physics C - Multiple Choice Practice Questions on Circular Motion and Rotation

“If I have seen a little further it is by standing on the shoulders of Giants.”
Sir Isaac Newton

A significant number of students in any class have the wrong notion that circular motion and rotation (or, generally angular motion) is a somewhat difficult section. But you can rest assured that this section is as interesting and simple as any other section of your choice, provided you have a clear understanding of the basic principles. You will find the useful formulae and many questions in this section,  posted earlier on this site, which can be accessed by clicking on the label, ‘circular motion and rotation’ below this post.  Since the number of posts displayed in each screen is limited, you will have to make use of the ‘older posts’ tabs to access all the posts.

Here are some more multiple choice practice questions (with solution) on circular motion and rotation:

(1) A thin uniform iron rod of mass M and length L is resting on a smooth horizontal surface. Two small objects each of mass m and traveling with uniform speed v in opposite directions at right angles to the length of the rod collide with the two ends of the rod simultaneously as shown in the adjoining figure. If the objects get stuck to the rod after the collision and continue to move along with the rod, the angular impulse received by the rod is

(a) mvL

(b) 2mvL

(c) mvL/2

(d) mvL/4

(e) mvL/12

The angular impulse received by the rod is equal to the change in the angular momentum of the rod. Since the rod is initially at rest, the change in the angular momentum of the rod is equal to the angular momentum supplied by the two objects.

Because of the collision, the rod will rotate about a normal axis through its middle and the total angular momentum of the objects about this axis of rotation is mvL/2 + mvL/2, which is equal to mvL.

Thus the angular impulse received by the rod is mvL.

(2) In the above question, if m = M/6 and v = L ms–1, then the time taken by the rod to rotate through π radian is

(a) 1 sec

(b) 2 sec

(c) π/2 sec

(d) π sec

(e) 2π sec

Because of the collision, the rod will rotate about a normal axis through its middle with an angular velocity ω given by

Iω = mvL where ‘I’ is the moment of inertia of the rod and the masses m and m at its ends.

[Note that we have equated the final angular momentum of the system (containing the rod and the masses) to the initial angular momentum].

Since v = L the above equation gets modified as

Iω = mL2

After the collision, the rod and the masses move together and the total angular momentum is given by

Iω = [(ML2/12) + 2m(L/2)2

[The first term within the square bracket above is the moment of inertia of the rod about a central axis perpendicular to its length and the second term is the moment of inertia of the two masses].

From the above two equations, we have

mL2 = [(ML2/12) + mL2/2

Since m = M/6 the above equation becomes.

M/6 = [(M/12) + (M/12)]ω = (M/6)ω

Therefore, the time taken by the rod to rotate through π radian is π sec.

(3) The angular momentum of the electron due to its orbital motion in the hydrogen atom is directed

(a) along the direction of motion of the electron.

(b) opposite to the direction of motion of the electron.

(e) normal to the plane of the orbit.

The correct option is (e). The angular momentum vector is the vector product (cross product) of the radius vector r and the linear momentum vector p (of the electron). The angular momentum vector r×p is perpendicular to both r and p. Therefore, the angular momentum of the electron due to its orbital motion must be directed normal to the plane of the orbit.

(4) A solid cylinder of mass M and radius R is mounted on frictionless bearings so that it can rotate about its horizontal axis (Fig.). A light inextensible string, one end of which is fixed to the cylinder, is wound round the cylinder and carries a sphere of mass M. Initially the system is at rest. On releasing the sphere, it moves down under gravity and the cylinder starts rotating. What is the speed of the sphere after it has descended through a distance h?

(a) √(2gh)
(b) √(4Rgh/3)
(c) √(4gh/3)
(d) √(2Rgh)
(e) √(4gh)

The gravitational potential energy lost by the sphere (on falling through the height h) is mgh. This can be equated to the kinetic energy gained by the cylinder and the sphere. Therefore we have

Mgh = ½ Mv2 + ½ 2 where v is the speed of the sphere on falling through the distance h, I is the moment of inertia of the cylinder about its axis and ω is the angular velocity of the cylinder.

[Note that the sphere has translational kinetic energy and the cylinder has rotational kinetic energy].

Since I = ½ MR2 and ω = v/R we can rewrite the above equation as

Mgh = ½ Mv2 + ½ (½ MR2)(v2/R2)

Or, gh = (v2/2) + (v2/4) = 3v2/4

This gives v = √(4gh/3)