REG_Q01=ADV_Q01

Semester 1, Y2011 Question 1

A large wooden turntable in the shape of a flat uniform disk is freely rotating about a vertical axis through its centre. A parachutist descends vertically and makes a soft landing on the turntable, without slipping, at a point near its outer rim.

(a) Will the angular speed of the turntable increase or decrease after the parachutist lands? Briefly justify your answer.

(b) Will the kinetic energy of the system (turntable + parachutist) increase or decrease in the process? Briefly explain your answer.

(5 marks)

Solution

(a) Angular momentum L I= will be conserved in the process. The addition of the parachutist will increase I and hence will decrease. (2 marks) (b) We have 1 1 2 2I I = where, 1I is the initial moment of inertia of the turntable and 1 is its angular velocity, 2I is the moment of inertia of the turntable plus parachutist and 2 is its angular velocity. Therefore

12 1 12

II

= < since 2 1.I I> The kinetic energy is

22 2 21 1

2 2 2 2 1 1 122 2

11

2

1 1 12 2 2

I IK I I II I

I KI

= = =

=.

So 2 1K K< since 2 1.I I> Hence kinetic energy decreases. Friction will do net negative work on the system (turntable + parachutist) in the process of accelerating the parachutist in the horizontal direction. The friction would in reality most likely arise from slipping between the parachutist and the turntable. But it could also arise from deformation of the parachutist as he/she is accelerated to the velocity of the turntable. Energy is lost as a result. (3 marks)

REG_Q02

Semester 1, Y2011 Question 2 A ball of mass m at the end of a light string is swung so that it moves in a vertical circle as shown in the diagram. The length of the string is R . At the top of its motion the ball has velocity .v

(a) Draw free-body diagrams showing all the forces acting on the ball when the ball is in

each of the positions

(i) A

(ii) B

(iii) C.

(b) What is the direction of the net force on the ball when it is in position C? Briefly explain your answer.

(c) What is the minimum velocity of the ball at position A necessary to keep the ball moving in a circle of radius R ?

(5 marks)

Solution (a)

( mark for T in each diagram, mark for weight force=2 marks) (b) Net force on ball at Position C must be upwards towards the centre of the circle in which the ball is being swung. This is needed to keep the ball moving in a circle. Hence:

.F T m g= (1 mark) (c) At position A

F T m g ma= + = where a is the radial acceleration towards the centre of the circle of radius R . If the ball is travelling in circular motion with velocity v at point A then we have

2vaR

= . Hence

2mvT m gR

+ = . The minimum value of v occurs when 0T = and so

2mvm g v g RR

= = .

( mark for force equation, mark for using 2vaR

= , mark for 0T = , mark for correct answer)

REG_Q03

Semester 1, Y2011 Question 3

Consider a solid disk with mass M and radius R (moment of inertia 21

2I M R= ) rolling

without slipping on a surface. The disk has a constant centre of mass velocity CoMv and angular speed . (a) Draw a free-body diagram showing the forces on the disk.

(b) Write down an equation relating CoMv and . Briefly explain your relation. (c) If transK is the kinetic energy of the disk due to its translational motion and rotK is the

kinetic energy due to its rotation, show that the ratio /trans rotK K is independent of the mass and radius of the disk. Briefly explain all steps in your derivation.

(5 marks)

Solution (a)

(1 mark) (b) Since the disk is rolling without slipping the contact point is instantaneously at rest with respect to the ground. The centre of mass speed is CoMv so that the point at the rim must have a linear speed CoMv R= for its velocity of rotation to be equal and opposite to its translational velocity when it is in the contact position. (2 marks) (c) The kinetic energy due to translation motion is

212tran CoM

K M v= The kinetic energy due to rotation is

212rot

K I=

where I is the moment of inertia for a solid disk 212

I M R = .

Using the relationship CoMvR

= we have 2

2 21 1 12 2 412

CoMrot CoM

tran

vK M R M vR

K

= = =

So / 2tran rotK K = which is independent of mass M and radius R . (2 marks)

REG_Q04

Semester 1, Y2011 Question 4 (a) Two mercury-in-glass thermometers initially at room temperature are placed in a large

mug which is nearly full of water at 80 CD . One thermometer has a large bulb and the other a small bulb (see diagram on left below).

The figure above on the right shows the temperature change of the two thermometers in the first minute.

(i) Which line (A, B, C, or D) best represents the temperature of the thermometer with the large bulb? Briefly explain your answer in terms of the physics of heat transfer and thermal equilibrium.

(ii) Would you use the thermometer with the large bulb to measure the temperature of a very small amount of water that just covers the bulb? Explain your answer.

(b) Three identical thermometers are outside and are initially at the surrounding air temperature. Thermometer 1 is placed in the shade, thermometer 2 is placed in the sun, and thermometer 3 is wrapped with black material and placed in the sun.

Describe the variation with time of the temperature measured by each thermometer and explain the reasons for any differences between the response of the three thermometers.

(5 marks)

Solution (a)(i) Line C is the most appropriate curve for the thermometer with the large bulb. The thermometers will rise over time from room temperature to the water temperature of 80 CD . Hence lines A and D are not correct.

( mark) The rate of heat transfer into the mercury is proportional to the surface area of the bulb. The rate of temperature rise proportional to the rate of heat transfer into the bulb but inversely proportional to the mass of the mercury in the bulb. In turn the mass of mercury in the bulb is proportional to the volume of the bulb. Heat transfer into bulb:

( )H CdQ k A T Tdt = ( mark) Rate of temperature rise:

1 1dQ dT dT dQ dQmCdt dt dt mc dt V c dt= = =

( mark) Rate of temperature rise is therefore proportional to:

( )2

3

4 14 / 3

dT A Rdt V R R

= .

So the thermometer with the largest radius has the slowest rise in temperature. Hence line C is the appropriate answer. ( mark) (a)(ii) No. If the bulb is large enough then thermal equilibrium is not reached. The thermometer influences and changes what it is intended to measure. (1 mark) (b) Thermometer 1 stays at air temperature. Heat transfer occurs by conduction and convection and the thermometer is already at equilibrium with these processes.

( mark) Thermometers 2 and 3 are not initially at equilibrium and will show a temperature rise because the heat transfer also includes radiation from the sun.

( mark)

They will both reach the same equilibrium temperature but Thermometer 3 will have the fastest temperature rise as a result of the black material having higher emissivity. This can be seen from the Stefan-Boltzmann law ( )4 4 ,sH A e T T= where sT is the temperature of the surroundings and T is the temperature of the thermometer. The higher value of emissivity e for Thermometer 3 means a higher value of H for the same values of temperature. The rate of temperature rise for Thermometer 3 will therefore be larger as seen from:

T Hmc T H tt mc

= = (1 mark)

REG_Q05=ADV_Q05

Semester 1, Y2011 Question 5

The diagram above shows a wire with its left end attached to an oscillating pin that is fixed to a table. The wire extends to the right over a frictionless pulley and is attached to a hanging weight of mass 5.00 kg . The distance between the pin and the pulley is 1.00 m and the linear density of the wire is 10.100kg.m . The wire is made to vibrate when the pin is connected to a frequency generator causing it to oscillate up and down. The oscillating pin generates a sinusoidal travelling transverse wave in the wire with a frequency of 40.0 Hz and amplitude of 0.020 m.

(a) Calculate the speed of the wave, neglecting the mass of the wire.

(b) Write down an equation that describes the vertical displacement of the wave y as a function of horizontal distance x from the pin and time t . Assume a travelling wave with no reflections. Insert numerical values for any constants in the equation.

(c) What is the lowest frequency of the vibrating pin that will generate a standing wave in the wire?

(5 marks)

Solution (a) The speed is given by

Fv = where 10.100kg.m = and ( )( )5.00 9.80 49.0 N.F mg= = = Therefore

149 22.1m.s .

0.10v = =

(2 marks) (b)

( ) ( ), cosy x t A k x t= For this situation we have 0.020 mA =

12 2 (40.0) 251.3 251rad.s .f = = = =

1251.3 11.4 rad.m .

22.1k

v = = =

And hence

( ) ( ), 0.020cos 11.4 251y x t x t= (student is free to use plus or minus in the cos argument as well as any additional phase/offset).

( mark for each of , ,A k and also mark for correct statement of ( , )y x t ) (c) The fundamental frequency is given by

( )( )22.1 11.1Hz

2 2 1.0vfL

= = = . (1 mark)

REG_Q06=FND_Q06

Semester 1, Y2011 Question 6 Helen and Greg are two instrument builders who are competing to see whose instrument can play the note with lowest frequency. Helen builds a giant saxophone of length 4.0 m, which acts like an open pipe. Greg builds a giant clarinet of length 4.0 m, which acts like a closed pipe. The speed of sound in air is 1344 m.s .

(a) Draw a sketch showing the fundamental mode for Helens saxophone.

(b) Draw a sketch showing the fundamental mode for Gregs clarinet.

(c) Which builder, Helen or Greg, wins the competition?

(d) Calculate the frequency of the lowest note for the winning instrument.

(5 marks)

Solution (a)

Helens Saxophone

(1 mark) (b)

Gregs Clarinet

These are displacement graphs. Students could also draw graphs of pressure and should be marked as correct. (1 mark) (c) Lowest note = longest wavelength (1 mark) Longest wavelength on Helens saxophone (open pipe) = 2 length 8.0 m = Longest wavelength on Gregs clarinet (closed pipe) = 4 length 16.0 m = Therefore Gregs clarinet is the winner of the competition.

(1 mark for winner with justification) (d)

Frequency 344 21.5Hz 21Hz.16

vf = = = = (approximately the lowest frequency the human ear can hear) (1 mark)

REG_Q07

Semester 1, Y2011

Question 7

A 70 kg skier starts at the top of a slope that is inclined at an angle of 12D to the horizontal. The slope is 200 m long and the coefficient of kinetic friction between the slope and the skis is 0.075.

(a) Draw a free-body diagram showing all the forces on the skier as he slides down the slope.

Assume that he does not have ski poles, just skis on his feet.

(b) Calculate his acceleration down the slope.

(c) Calculate his speed when he reaches the end of the slope.

When he takes the ski tow back up the slope, the cable pulls him at a constant speed. The cable is at an angle of 25D to the slope.

(d) Draw a free-body diagram of the forces now acting on the skier. Assume that there is the

same coefficient of kinetic friction between the skis and the slope.

(e) What is the tension in the cable as the skier is pulled up the slope?

(10 marks)

Solution (a)

(1 mark)

For problem 12 = D , 70 kgm = . (b) Resolve forces in the x direction: sinxF m g f= . This is equal to ma where a is the acceleration down the slope: sin .ma m g f= (1) (1 mark) Resolve forces in the y direction: cos .yF n m g = This is equal to zero as there is no acceleration in the y direction: 0 cos .n m g = (2) (1 mark) The frictional force is ( )0.075 .k kf n = = From equation (2) cos .kf m g = Substitute this result into equation (1):

( )( ) ( )( )( )2

sin cossin cos9.80 0.208 0.075 9.80 0.978

1.319 (3 sig figs 1.32) m.s .

k

k

ma m g m ga g g

= = = =

(1 mark)

(c) The speed at the bottom of the slope is:

( )2 20 02 0; 1.319; 200 .v v a s v a s= = = = So

( )( )( )2

1

2 1.319 200 528

23.0 (23)m.s .

v

v = = =

(2 marks) (d)

(1 mark) For diagram 12 = D , 70 kgm = , 25 = D . The skier is moving upwards at constant speed. Note the the coordinate system now has upwards along the slope as the positive x direction. (e) Resolve forces in the x direction: cos sinxF T m g f = . This is equal to ma where 0a = because the skier is moving at constant speed along the slope. 0 cos sin .T m g f = (3) (1 mark) Resolve forces in the y direction: sin cos .yF T n m g = + This is equal to zero as there is no acceleration in the y direction: 0 sin cos .T n m g = + (4) (1 mark) The frictional force is again ( )0.075 .k kf n = = Multiply equation (4) by k to get:

0 sin cos .k k kT n m g = + (5)

Add equation (3) and (5) together to get:

( ) ( ) ( )cos sin cos sin 0.k k kT T n f mg + + + = And therefore:

( ) ( )( )( )

( )( ) ( )( ) ( )( ) ( )( )

cos sin cos sin

cos sincos sin

0.075 0.978 0.20870 9.8

0.906 0.075 0.423206 N.

k k

k

k

T T mg

T mg

+ = ++= +

+= +=

(1 mark)

FND_Q09=REG_Q08

Semester 1, Y2011 Question 9 or 8 A 0.060 kg arrow is moving vertically upward at a speed of 150 m.s just as it hits a block of wood of mass 5.0 kg . The arrow stops in the block and the block is free to move vertically.

(a) What is the kinetic energy of the arrow just before the collision?

(b) What is the momentum of the arrow just before the collision?

(c) What is the momentum of the block (with arrow embedded in it) just after the collision?

(d) What is the velocity of the block just after the collision?

(e) How much energy was lost in the collision?

(f) How high does the block rise above its initial position?

(10 marks)

Solution (a)

Just before the collision the kinetic energy is:

2 21 (0.5)(0.06)(50 ) 75 J.2

K m v= = =

(2 marks) Answers to parts (b), (c) and (d) are vectors. Deduct 1 mark if directions are not given (b) Just before the collision the magnitude of the momentum is:

1(0.06)(50) 3.0 kg.m.s .p mv = = = The momentum is directed upwards.

(2 marks) (c) Since momentum is conserved, the combined momentum of the block + arrow after the collision is

13.0 kg.m.s .ip p= =

The momentum is directed upwards. (1 mark) (d) The magnitude of the momentum after the collision is:

( ) 11

3.0 kg.m.s

3.0 0.59 m.s .(5.0 0.06)

f

f

p M m v

v

= + = = =+

The velocity is directed upwards. (1 mark) (e) The kinetic energy just after the collision is

2 21 ( ) (0.5)(5.06)(0.59 ) 0.88 J,2 f

K m M v= + = = (or 0.89 J if all the sig figures are kept for fv ). Hence the energy lost is:

75 0.88 74 JE = = (i.e. nearly all of the energy). (1 mark for method; 1 mark for correct result) (f) The block will rise until all of the kinetic energy is transformed into potential energy:

( )0.88 (5.06)(9.81) ,0.88 0.018m 1.8cm.

(5.06)(9.81)

K J m M g h h

h

= = + = = = =

(1 mark for method; 1 mark for correct result)

REG_Q09=ADV_Q09

Semester 1, Y2011 Question 9 A sample of 2.00 mol of Helium gas (assumed to be ideal, monatomic gas) has a volume of

30.0326m , a pressure of 1.50 atm , and a temperature of 25.0 CD (State A). A physicist firstly heats the gas at constant volume, adding 41.50 10 J of heat, to reach State B. He then continues heating and allows the gas to expand at constant pressure to twice its original volume, reaching State C. Answer the following questions, in each case making it clear how you obtained your results in terms of known parameters such as , , etcA A Ap V T .

(a) Sketch a pV diagram for the whole process showing the positions of States A, B and C. For now, you need not calculate any unknown values for p and V .

(b) Write down an expression for the molar heat capacity of the gas at constant volume and use it to calculate the temperature of the gas at State B.

(c) Calculate the pressure and the temperature of the gas at State C.

(d) Calculate the amount of work done by the gas.

(e) Calculate the change in internal energy of the gas for the whole process from A to C.

(10 marks)

Solution Pressure at State A is ( )( )5 51.50 1.013 10 1.52 10 Pa. = Temperature at State A is 273.15 25.0 298.2 K+ = The table below shows the values (or relationships) given in the problem and those calculated later in the solution (highlighted text). Pressure Volume Temperature State A 51.520 10 Pa 30.0326 m 298.2 K State B 54.59 10 Pa Same as A 30.0326 m 899.6 K

State C Same as B

54.59 10 Pa Double B

30.0652 m 1800 K

(a) The processes are shown in the following pV diagram. The actual values for p and V are not required.

(1 mark) (b) Molar heat capacity for an ideal monatomic gas at constant volume is given by

3 .2V

C R= (1 mark)

( )( )( )41.50 10 601.4 K

2.00 1.5 8.314V V

dQdQ nC dT dTnC

= = = =

So temperature at State B is ( )601.4 273.15 25.0 899.6K+ + = (2 marks) (c) Calculate the pressure at State B Bp from the ideal gas law as follows:

( )( )( ) ( )5 52.00 8.314 899.6 4.589 10 4.59 10 Pa.0.0326

BB B B B

B

n RTp V n RT pV

= =

= =

The change from State B to State C is at constant pressure and so

54.59 10 Pa.C Bp p= =

(1 mark) Calculate the temperature at State C from the ideal gas law as follows:

( )( )( )( )( ) ( )

54.589 10 2 0.03261799 1800 K.

2.00 8.314

C CC C C C

p Vp V n RT Tn R

= == =

(1 mark) (d) Work done by the system is given by W p dV= We can consider the two processes separately as follows: State A State B This is isochoric (constant volume) and there is no work done by the gas as 0.dV = (1 mark) State B State B This is isobaric (constant pressure) and the work done bt the gas can be written

( )

( )( ) ( )5 44.589 10 0.0326 1.496 1.50 10 J.C C

B B

V V

B B C B B AV V

W p dV p dV p V V p V= = = =

= =

(1 mark) (e) Internal energy of an ideal gas is determined only by the change in temperature.

( ) ( )( )( )( )4

3 1.5 2.00 8.314 1799 298.22

3.74 10 J.

V C AdU nC dT n R T T= = = = (2 marks)

REG_Q10

Semester 1, Y2011 Question 10 Two identical slightly curved metal Magdeburg plates are placed rim to rim with airtight seals (as in the diagram) and the interior region is pumped down to a vacuum. The area of each plate is

21.5m . This assembly is initially placed in air at 400 K and at atmospheric pressure of 51.0 10 Pa .

(a) Explain qualitatively at a microscopic level (kinetic theory) why it is difficult to pull the

plates apart.

(b) Calculate the force necessary to pull the two plates apart, approximating the curved plates as being flat.

A small puncture is then made through the seal and air at 400 K is allowed to expand rapidly (adiabatically) into the region between the plates. The expanding air cools and then the metal plates cool.

(c) Explain physically why the expanding air cools and then the metal plates cool, referring to the changes in internal energy, work, and heat for the (ideal) gas.

(d) Explain why the entropy change of the expanding gas is zero.

(e) Consider the energy interchange between the air and metal plates as they come to equilibrium. Is the total entropy change greater than, equal to, or less than zero? Justify your answer by calculating the entropy change for an infinitesimal amount of heat dQ between the initially hotter plates and the air.

The entire assembly is now heated to 600 K . The thermal radiation from the assembly is then observed and compared with the radiation observed when the temperature is 400 K .

(f) By what factor does the overall energy flux of thermal radiation differ when 600 KT =rather than 400 K.T = . Assume that the assembly radiates as a blackbody.

(10 marks)

Solution (a) The external surfaces are subject to collisions with air molecules, with each collision involving a change in momentum and so a net force exerted inwards. In the vacuum region there are no air molecules and so no compensating externally-directed force due to collisions. Accordingly, the external collisions exert a net force and so pressure (= force/unit area) directed inwards. Pulling the plates apart requires exertion of a net force greater than the inwards-directed force exerted by air molecules via collisions. (2 marks) (b) Required force is atmospheric pressure multiplied by the area of the plates. This is given by ( )( )5 51.0 10 1.5 3.0 10 NaF p A= = = (1 mark) (c) The gas expands rapidly (adiabatically) and so there is no heat transfer by definition ( )0dQ =( mark) Work is done by the gas as it expands, since dW p dV= and the volume increases ( mark). Accordingly the change in internal energy 0dU dQ dW= < . ( mark) For an ideal gas U is a function only of temperature (for constant mass) with U increasing with increasing T . Accordingly the gas cools. ( mark) (d) The change in entropy /dS dQ T= . Accordingly the entropy change of the gas is zero as 0dQ = (adiabatic expansion) as it expands into the vacuum. (1 mark) (e) The entropy change for the gas-plate system coming to equilibrium is greater than zero. (1 mark) Since the metal is originally at a higher temperature mT then its change in entropy as it transfers

0dQ > to the gas is / mdQ T . The entropy change for the gas is then / gdQ T with g mT T< . The total entropy change is then 1 1 0

g m

dQT T

> since g mT T< .

(2 marks) [Partial credit could be given (1 mark) for the justification that the entropy change is greater than zero by the Second Law, since this transition to equilibrium is not a reversible process.]

(f) By the Stefan-Boltzmann Law the radiated power is proportional to 4T . Accordingly the radiated

power will increase by a factor of ( )4 4600 1.5 5.1 (5).400

= = (1 mark)

REG_Q11

Semester 1, Y2011 Question 11 A block is placed on a frictionless surface and attached to a horizontal spring of negligible mass with a force constant of 210.0 kg.sk = . At time t = 0, the block is displaced 0.200 m from equilibrium and is observed to oscillate in simple harmonic motion with a period of 8.00s .

(a) What is the velocity of the block 2.00 seconds after it is released? (Hint: Consider the

energy of the oscillating system)

The figure below shows the displacement of the block (relative to its equilibrium) versus time.

(b) Copy the figure into your answer book. Insert and label axes on the figure with

appropriate numerical values.

(c) On your figure, mark clearly the position(s) where the horizontal acceleration of the block is zero.

(d) On your figure, mark clearly the position(s) where

(i) the potential energy of the block is maximum, and

(ii) the position(s) where the kinetic energy of the block is maximum.

(e) The block is now placed on a rough surface that produces a frictional force that is proportional to the speed of the block. The constant of proportionality is 130.0 kg.sb = . The block is displaced from equilibrium and released. Describe the motion of the block.

(10 marks)

Solution (a) The period of the SHM is 8.00sT = . The motion starts from a point of maximum displacement at 0t = and thus at 2.00st = an interval of / 4T has elapsed and the mass must have returned to its equilibrium point at 0x = . (1 mark) The mass of the block can be calculated from the period of the oscillation from the equation.

2

22 4m kTT mk

= = . For this case we have

( )( ) ( )2

2

10.0 8.0016.21 16.2 kg.

4m = = (1 mark) At its initial point the velocity is zero and there is no kinetic energy. All of energy is stored as elastic potential energy. This is given by

21 .2

U k A= where 0.200 mA = is the amplitude of the oscillation. At the equilibrium point there is no stored elastic potential energy and all of the energy is in the form of kinetic energy. This is given by

212

K mv= . The initial and the final energies must be equal and hence

2 21 12 2

K mv U k A

kv Am

= = =

=

For this case we have

( ) 110.00.200 0.157m.s .16.21

v = = (1 mark) Note that you do not need to explicitly solve for the mass of the block but can work through using just the period of the oscillation. (b) See graph below

(1 mark for correct horizontal scales and labels) (1 mark for correct horizontal scales and labels) (c) See diagram below

(1 mark)

(d) See graph below

(1 mark for potential energy) (1 mark for kinetic energy) (e) For critical damping

( )( ) ( )1 1 110.0 8.002 25.5kg.s N.m .s .critical k Tb k m = = = = (1 mark) The value of b for this surface is larger than the critical value and so the block will return to equilibrium without oscillating. (1 mark)

REG_Q12=FND_Q12=ADV_Q12

Semester 1, Y2011 Question 12 Richard wishes to test the Doppler effect for sound waves. To do so, he attaches a sound wave generator to the arm of a large centrifuge (normally used to train astronauts and pilots to withstand high g forces) as shown in the diagram below. The distance between the centre of rotation and the sound generator is 10.0 m . The sound generator produces a single tone of frequency 2000 Hz and the centrifuge is set to make a complete revolution once every 2.0s . The speed of sound in air is 1344 m.s . A sound detector is placed at rest at a large horizontal distance (much larger than the size of the centrifuge) to the right of the centrifuge and is used to measure the frequency and wavelength of the sound produced by the generator.

(a) Calculate the tangential speed of the sound generator as the centrifuge rotates.

(b) Calculate the wavelength of the sound measured by the detector before the centrifuge begins to rotate.

(c) The detector finds that the frequency of the sound wave is changing as the centrifuge rotates. Explain why this occurs.

(d) Describe how the detected sound frequency changes as the centrifuge rotates, indicating at which point(s) (numbered 1, 2, 3, 4 in the above diagram) the frequency is highest and at which point(s) it is lowest.

(e) What is the value of the highest frequency measured by the detector?

(f) What is the wavelength of the sound measured by the detector when the frequency is highest?

(10 marks)

Solution (a) Circumference of the circular path of the sound generator is:

( )2 2 10.0 62.83m.C R = = = This is covered in one rotation period of 2.0s.T = Hence the tangential speed of the sound generator is:

( )1 12 31.42m.s 2 sig figs 31m.st Rv T = = . (1 mark) (b) Frequency of the sound at rest is 2000 Hz. Wavelength is calculated from

( )344 m 0.172m 2 sig figs 0.17m .2000

vv ff

= = = = (1 mark) (c) The velocity of the generator relative to the detector is constantly changing, so that the frequency of the sound is Doppler-shifted, and this shift is constantly changing as the generator rotates. (2 marks) (d) The frequency observed is highest when the generator is approaching the detector (position 3) and lowest when the generator is moving directly away from the detector (position 1). (1 mark for explanations, 1 mark for positions) (e) The highest frequency is observed when the generator is moving towards the detector at position 3. At this time the velocity of the generator towards the detector is that given in part (a).

LL S

S

v vf fv v+=

where v is the velocity of sound, Lv is the velocity of the listener (the detector) relative to the medium (air), and Sv is the velocity of the source relative to the medium (air), and sf is the frequency of the source. For this situation we have

10 m.sLv=

131.42 m.sSv= +

1344m.sv = Therefore

( ) ( )344 02000 Hz 2200 Hz.

344 31.42Lf += =

(2 marks)

(f) Wavelength is given by

( ) ( )344 31.42 0.156 m 2 sig figs 0.16m .2000

S

S

v vf

= = = . Alternatively

344 m 0.16 m.2200L

vf

= = = (2 marks)

REG_Q01=ADV_Q01_SolutionREG_Q02_SolutionREG_Q03_SolutionREG_Q04_SolutionREG_Q05=ADV_Q05_SolutionREG_Q06=FND_Q06_SolutionREG_Q07_SolutionFND_Q09=REG_Q08_SolutionREG_Q09=ADV_Q09_SolutionREG_Q10_SolutionREG_Q11_SolutionREG_Q12=FND_Q12=ADV_Q12_Solution