AP Physics B - Wave Motion (including sound) - Multiple Choice Practice Questions on Sound

Questions on sound were discussed earlier on this site. You can access all posts related sound by clicking on the label ‘sound’ below this post.

Today we will discuss a few more multiple choice practice questions on sound, relevant to AP Physics B aspirants.

(1) A source producing sound of wave length 0.6 m is moving away from a stationary listener with speed V/6 where V is the speed of sound in air. The wave length of sound heard by the listener is

(a) 0.5 m

(b) 0.54 m

(c) 0.66 m

(d) 0.7m

(e) 0.8 m

You might have come across questions of the above type under Doppler effect.

If the real frequency of the source of sound is n we have

V/n = λ where λ is the wavelength of the sound as measured by the stationary listener when the source is stationary. Therefore we have

V/n = 0.6 ……………..(i)

When the source moves away from the listener at speed V/6 the n sound waves produced per second will occupy a distance V + V/6 so that the wave length as measured by the listener becomes (V + V/6)/n. Therefore we have

(V + V/6)/n = λ₁ ……….(ii)

where λ₁ is the new wave length.

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

1 + 1/6 = λ₁/0.6

Or, 7/6 = λ₁/0.6 from which λ₁ = 0.7 m.

(2) Two steel wires A and B of the same length are vibrating under the same tension. If the first overtone of wire A has the same frequency as the fundamental note of wire B, the area of cross section of wire A is

(a) twice that of B

(b) three times that of B

(c) four times that of B

(d) half that of B

(e) one third that of B

The frequency of vibration (f) of a string of length ℓ and linear density m stretched under tension T is given by

f = (s/2ℓ) √(T/m) where s = 1,2,3,4 etc. For the fundamental mode s = 1. For the 2^nd mode (or 1^st overtone) s = 2. For the 3^rd mode (or 2^nd overtone) s = 3 and so on.

The linear density is mass per unit length and is equal to aρ where ‘a’ is the area of cross section and ρ is the density of the material of the string.

Since the first overtone of wire A has the same frequency as the fundamental note of wire B, we have

(2/2ℓ) √(T/m₁) = (1/2ℓ) √(T/m₂) where m₁ is the linear density of A and m₂ is the linear density of B.

From mthe above equation it follows that m₁ = 4 m₂.

Since the wires are of the same material, the area of cross section of A must be 4 times that of B.

[Since the changes are confined to the mode of vibration and the linear density, you should be able to write 2/√m₁ = 1/√m₂ to arrive at m₁ = 4 m₂ and the final answer a₁ = 4 a₂]

(3) Two sound notes of wave lengths λ₁ and λ₂ in air produce n beats per second. If λ₁ < λ₂ the speed of sound in air is

(a) λ₁λ₂n /(λ₂ + λ₁)

(b) (λ₁ + λ₂)n

(c) (λ₁λ₂ /n )(λ₂ + λ₁)

(d) λ₁λ₂n / λ₁

(e) λ₁λ₂n /(λ₂ – λ₁)

The frequency f₁ of the sound of wave length λ₁ is given by

f₁ = v/λ₁ where v is the speed of sound in air.

The frequency f₂ of the sound of wave length λ₂ is given by

f₂ = v/λ₂

Since λ₁ < λ₂ we have f₁ > f₂

Since the number of beats produced per second is n, we have

f₁ – f₂ = n

Therefore, v/λ₁ – v/λ₂ = n

Or, v (1/λ₁ – 1/λ₂) = n

This gives v = λ₁λ₂n /(λ₂ – λ₁)

(4) A glass tube is open at both ends. The minimum frequency of a tuning fork which vibrates in resonance with the air column in this tube is f. If this tube is held vertically with half its length submerged in water, what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube?

(a) f/3

(b) f/2

(c) f

(d) 2f

(e) 3f

The air column will vibrate with the minimum frequency in the fundamental mode. When both ends of the glass tube are open, there are anti-nodes at the ends and the adjacent node in the middle so that the length L of the tube is equal to λ/2 where λ is the wave length of the fundamental note. Thus we have

L = λ/2

Or λ = 2L

[Remember that the distance between consecutive anti-nodes (or, consecutive nodes) is equal to λ/2].

When half the length of the glass tube is immersed in water, it becomes a closed pipe of length L/2. In the fundamental mode of vibration of the air column in the tube in this case there is a node at the closed end (water surface) and the adjacent anti-node at the open end as shown in the figure. Therefore we have

L/2 = λ₁ /4 where λ₁ is the wave length of the resonating sound.

Or λ₁ = 2L

In both cases the wave length of the resonating sound is the same. Therefore the resonant frequency is the same as f [Option (c)].

(5) In the above question what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube if a quarter of the tube is submerged in water?

(a) 2f/3

(b) 3f/4

(c) f

(d) 2f

(e) 3f

Initially, when both ends are open, the resonating wavelength as shown above is given by

λ = 2L

When a quarter of the glass tube is submerged in water, we have

3L/4 = λ₁/4

Therefore λ₁ = 3L

From the above equations λ/λ₁ = 2/3

Since the frequency is inversely proportional to the wave length we have

λ/λ₁ = f₁/f

Therefore f₁/f = 2/3 from which f₁ = 2f/3

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

“Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time”.

– Thomas A. Edison

Questions on circular motion and rotation were discussed earlier on this site. You can access them by clicking on the label ‘circular motion and rotation’ below this post’. Today we will discuss a few more interesting multiple choice practice questions in this section:

(1) A long, flexible ribbon of mass M rolled in the form of a cylinder (Fig.) of radius R is placed on horizontal floor. When a negligibly small push is applied, the ribbon unrolls without sliding along the surface (thereby reducing the radius of the cylinder). The horizontal velocity of the axis of the cylinder when the radius becomes (¾)R is

(a) √(2Rg)

(b) √(3Rg/2)

(c) √(27Rg/17)

(d) √(37Rg/27)

(e) √(47Rg/37)

When the roll of ribbon unrolls, the radius of the cylindrical roll decreases, thereby lowering the centre of gravity of the cylinder. Initially the mass of the cylinder is M and its centre of gravity is at a height R. When the radius of the cylinder becomes (¾)R, the mass of the cylinder is (9/16) M and its centre of gravity is at a height (¾)R

[When the radius is R, mass = M = πR²Lρ where L is the length and ρ is the density of the cylinder. When the radius is (¾)R, mass = π(9/16)R²Lρ = (9/16) M].

The initial gravitational potential energy of the cylinder is MgR and the final gravitational potential energy is (9/16)Mg×(¾)R = (27/64)MgR.

Therefore the loss of potential energy = MgR – (27/64)MgR = (37/64)MgR

The cylinder gains an equal amount of kinetic energy (partially rotational and partially translational) which is ½ Iω² + ½ mv² where I is the moment of inertia [of the cylinder of radius (¾)R] about its axis, ω is its angular velocity, m is its mass and v is its linear velocity. Thus we have

(37/64)MgR = (½)(mr²/2)(v²/r² )+ ½ mv² = (¾)mv²

[Remember that the moment of inertia of a cylinder of mas m and radius r about its own axis is mr²/2 and the angular velocity is v/r where v is the linear velocity].

Therefore, (37/64)MgR = (¾)(9/16)Mv²

This gives v = √(37Rg/27), as given in option (d).

[The above question is perhaps more suitable as part of a free response question].

(2) The angular momentum of a fly-wheel changes from 2L to 5L in 5 seconds under the action of a constant torque. The torque is

(a) 5L

(b) 3L

(c) 2L

(d) (3/5)L

(e) (2/5)L

The change in the angular momentum of the fly-wheel is 3L. The angular impulse received by the fly-wheel during the time t is τ t where τ is the torque. Since the angular impulse is equal to the change in angular momentum, we have

τ t = 3L

Since t = 5 seconds, τ = (3/5)L

(3) The adjoining figure shows a bar pendulum. It consists of a sphere of mass M attached to the end B of a light rigid rod AB suspended from a horizontal nail (at A) on the wall. Initially the pendulum is at rest in its equilibrium position, with the rod vertical. What is the magnitude of the minimum velocity to be imparted to the sphere so that it just completes its circular path?

(a) √(gL)

(b) √(2gL)

(c) √3gL)

(d) √(4gL)

(e) √(5gL)

Since the rod is light, this problem is quite simple. It is enough to equate the initial kinetic energy of the sphere to its potential energy (gravitational) at its topmost position. Therefore we have

½ Mv² = Mg×2L where ‘v’ is the magnitude of the minimum velocity to be imparted to the sphere.

[Note that the height through which the centre of gravity of the sphere rises is 2L].

Therefore, v = √(4gL)

[If the sphere is suspended by means of a string (as in the case of a simple pendulum), the minimum speed to be imparted will be greater {equal to √(5gL)}, since the sphere should possess kinetic energy also at its topmost position, in order to maintain its circular path. In that case the tension in the string is zero and the speed at the top of the path (v_top) is such that the centrifugal force on the sphere has the same magnitude as the weight of the sphere:

Mv_top²/L = Mg from which v_top = √(gL)

The kinetic energy at the top is ½ Mv_top² = ½ MgL

The minimum kinetic energy to be imparted to the sphere in its initial position is ½ MgL + (Mg×2L) = (5/2) MgL

Therfore, the minimum velocity ‘v’ to be imparted to the sphere is given by

½ Mv² = (5/2) MgL from which v = √(5gL)]

(4) A simple pendulum of length L has a bob of mass M. The pendulum is suspended from the point O and is held initially in the position OA, as shown in the adjoining figure, so that the string is horizontal. On releasing the pendulum, what will be the resultant acceleration of the bob when the pendulum is in the position OB? (Acceleration due to gravity = g)

(a) g [1 + 3 sin²θ]^1/2

(b) g [1 + 3 cos²θ]^1/2

(c) 3g sin²θ

(d) 3g (1+sinθ)

(e) 3g/2

The forces acting on the bob of the pendulum in the position OB are the weight Mg of the bob acting vertically downwards and the tension T (Fig.) in the string which supplies the centripetal force required for the circular motion of the bob.

The speed v of the bob in the position B is given by

½ Mv² = Mg Lsinθ

Therefore v = √(2gL sinθ) ……….(i)

[In position B the bob has fallen down by a distance L sinθ, losing gravitational potential energy mg Lsinθ and gaining kinetic energy ½ Mv²]

The centripetal force on the bob is the resultant of the tension T in the string and the component Mg sinθ of the weight of the bob. Therefore we have

Mv²/L = T – Mg sinθ

Substituting for v from Eq.(i) we have

2Mg sinθ = T – Mg sinθ

This gives T = 3Mg sinθ

The resultant force F acting on the bob is the resultant of the forces Mg and T (with an angle 90º+θ between them) and has magnitude equal to [(Mg)² + T² + (2×Mg× T) cos(90+θ)]^1/2.

Substituting for T, the resultant force is

F = [(Mg)² + (3Mg sinθ)² + (2× Mg× 3Mg sinθ) cos(90+ θ)]^1/2 = [(Mg)² + 3(Mg sinθ)²]^1/2 = Mg [1 + 3 sin²θ]^1/2

[Remember that cos(90+ θ) = – sinθ]

The resultant acceleration of the bob in position B is F/M = g [1 + 3 sin²θ]^1/2.

(5) A disc of mass M and radius R and a ring of mass M and radius 2R are released from rest from the top of an inclined plane. If they roll down the plane without slipping, which one of the following statements is correct?

(a) The ring will reach the bottom of the incline earlier

(b) They will reach the bottom of the incline with the same speed

(c) The ring will reach the bottom of the incline with twice the speed of the disc

(d) They will have the same acceleration down the incline

(e) They will reach the bottom of the incline with the same kinetic energy

The acceleration ‘a’ of a body of radius R rolling down an incline of angle θ is given by

a = g sinθ/[1 + k²/R²] where k is the radius of gyration of the body.

[The radius of gyration k is defined by I = Mk² where I is the moment of inertia (about the axis of rolling) and M is the mass].

Note that the acceleration is independent of the mass and size of the body. The ratio k²/R² also is independent of the mass and size of the body

For a given inclined plane, the acceleration depends only on the ratio k²/R².

For a given inclined plane, the acceleration depends only on the ratio k²/R².

[You may substitute the value of moment of inertia in the equation I = Mk² and find the value of k²/R². The maximum value in the case of regularly shaped bodies is for a thin ring and is equal to 1 since I = MR² = Mk². The least value is 2/5 which is the case for a solid sphere since I = (2/5)MR². Better, see this post ]

Moment of inertia of a disc (about the axis of rolling) is MR²/2 where as the moment pf inertia of a ring (about the axis of rolling) is MR².

The acceleration in the case of a disc is g sinθ/[1+ ½] = (2/3) g sinθ

The acceleration in the case of a ring is g sinθ/[1+ 1] = (½) g sinθ

All options except the last one (e) in the question are wrong. The kinetic energy with which the disc and the ring will reach the bottom of the incline will be the same since they have the same mass and they start from the same height.

[The kinetic energy at the bottom of the incline is the same as the potential energy at the top of the incline].