Combined IB HL Waves Qs 0809 MS

– 4 – N09/4/PHYSI/HP2/ENG/TZ0/XX/M+

SECTION A

A1. (a) both error bars of 15ms drawn correctly; [1]

(b) a straight line cannot be drawn through the

error bars; Accept the error bar comment with a

straight line drawn on graph.

that goes through the origin; [2]

(c) (i) 2 2500(m s ) ; [1]

(ii)

2

22

v v

v v;

2 2 527 2

27v ;

2 2 2( )300(m s )v or 2 2( )270(m s ) ; [3] or percentage error/uncertainty in (18.5 )19%v ;

percentage of error/uncertainty in 2 37 %v ;

absolute error2 2( )300(m s ) or 2 2( )270(m s ) ;

Answer must be to one or two significant figures.

(d)

2 3 2 2/ 10 m sv

3.0

2.5

2.0

1.5

1.0

0.5

0

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

T / N use of gradient triangle over at least half of line;

gradient 640( 40) ;

2k to give 640 25( 1)k ;

unit of k is 1 12 2kg m or

121ms N ; [4]

Do not penalize omission of factor of 1000 for missing y-axis label if already

penalized in (c). Treat as ecf.

2 3 2 22.3 10 m sv

( ) 3.6 NT

– 3 – SPEC/4/PHYSI/HP2/ENG/TZ0/XX/M

A2. (a) Intensity

Distance along screen

general shape; relative position of secondary maxima / relative heights of secondary maxima; [2] Award [1 max] if not touching x-axis. (b) A B

maximum of B coincides with first minimum of A; [1]

(c) 7

62

1.22 1.22 5 10 2.4 10 rad25 10b

λθ−

−−

× ×= = = ××

;

168.1 10x=

×;

to give 112.0 10 mx = × ; [3] A3. (a) polarized light is light in which the (electric) field vector vibrates in one plane only /

OWTTE; the liquid crystal changes the plane in which (electric) field vector rotates; [2] (b) (i) nothing / whole area black; since the optical axes of 1P and 2P are at right angles / OWTTE; [2] (ii) since the liquid crystal rotates the plane of polarisation light is now

transmitted by 2P / OWTTE; the electric field across the parts of the liquid crystal in the shape of the

electrode on G no longer rotates the plane of polarization; the field of view of the observer will now contain a black area

corresponding to the shape of the electrode on G / OWTTE; [3]


A2. (a) Intensity

Distance along screen

general shape; relative position of secondary maxima / relative heights of secondary maxima; [2] Award [1 max] if not touching x-axis. (b) A B

maximum of B coincides with first minimum of A; [1]

(c) 7

62

1.22 1.22 5 10 2.4 10 rad25 10b

λθ−

−−

× ×= = = ××

;

168.1 10x=

×;

to give 112.0 10 mx = × ; [3] A3. (a) polarized light is light in which the (electric) field vector vibrates in one plane only /

OWTTE; the liquid crystal changes the plane in which (electric) field vector rotates; [2] (b) (i) nothing / whole area black; since the optical axes of 1P and 2P are at right angles / OWTTE; [2] (ii) since the liquid crystal rotates the plane of polarisation light is now

transmitted by 2P / OWTTE; the electric field across the parts of the liquid crystal in the shape of the

electrode on G no longer rotates the plane of polarization; the field of view of the observer will now contain a black area

corresponding to the shape of the electrode on G / OWTTE; [3]


SECTION B B1. Part 1 Simple harmonic motion and the greenhouse effect (a) the force acting/accelerating (on the body) is directed towards equilibrium (position); and is proportional to its/the bodies displacement from equilibrium; [2] (b) (i) 101.5 10 m−× ; [1] (ii) 121.1 10 sT −= × ;

12

11.1 10

f −⎛ ⎞= ⎜ ⎟×⎝ ⎠

;

139.1 10 Hz= × [2] (iii) 14 1(2 ) 5.7 10 (rad );fω −= π = × ( )2 2 27 2 20 2 281 1

max 02 2 1.7 10 (1.5) 10 (5.7) 10E m xω − −= = × × × × × × ;

186.2 10 J−= × [2] (c) (i) 2 2 26 27

p(4 ) 40 83 10 1.7 10k f m −= π = × × × × ;

1560 N m−≈ [1] (ii) use of F = kx and F = ma;

to give 10

20 227

560 1.5 10 5.0 10 ms1.7 10

a−

−−

× ×= = ××

; [2]

(d) (i) infra red radiation radiated from Earth will be absorbed by greenhouse gases; and so increase the temperature of the atmosphere/Earth; [2] (ii) the natural frequency of oscillation (of a methane molecule) is equal to

139.1 10 Hz× ; because of resonance the molecule will readily absorb radiation of this

frequency; [2]


B3. Part 1 Simple harmonic motion

(a) 1. acceleration proportional to displacement from equilibrium/centre (of motion)

/mean position;

2. acceleration directed to equilibrium/centre/mean position; [2]

(b) (i) 2

d; [1]

(ii) sine/cosine curve shape reasonable; [1]

Do not allow semi-circle for half sine curve.

(iii) period labelled;

amplitude labelled; [2]

(c) (i) v 2a f seen/used;

13.3ms ; [2]

(ii) acceleration 2 24a f seen/used;

3 29.2 10 ms ; [2]

(d) cosine with the same period;

negative cosine;

Accept any amplitude. velocity

time

[2] Amplitudes need not be the same.

(e) (i) (a situation in which) a (resistive) force opposes the motion / the amplitude

decays with time; [1]

(ii) energy lost to surroundings / air resistance / frictional force is acting on the

fork; [1]

– 14 – M09/4/PHYSI/HP2/ENG/TZ1/XX/M+

B3. Part 1 Simple harmonic motion and waves

(a) is proportional to the displacement/distance (of the particle) from its equilibrium

position;

is directed towards the equilibrium position; [2]

(b) (i) overall correct shape;

with max of 0.06 J at x and zero at x ; [2]

(ii) max 0

2 2 2

K 4E mf x

;

from the graph maxK JE ;

and 0 mx ;

maxK

2

04

Ef

mx

;

to give Hzf [4] or

maxK

2

0

;E

kx

2

0.06;

0.05

= 48;

use of k

fm

;

= 2.0 Hz

(c) (i) the energy of the wave is propagated in a direction at right angles;

to the direction of oscillation of the particles; [2]

(ii) m ; [1]


(d) (i) use of 22 1

1

sin sinv

v ;

2

1

v

v ;

this marking point is not necessary to award full credit.

1

2 sin 0.75 49 ; [3]

(ii)

any two lines as shown bending in the correct direction; [1]


B3. Part 1 Simple harmonic motion and waves

(a) displacement is proportional to acceleration / vice versa;

because graph is straight-line through origin;

displacement and acceleration in opposite directions / acceleration always directed

towards origin;

because negative gradient; [4]

(b) use of 2 ( )a

x ;

2

3

2900

0.60 10

;

2 f ;

3

1 2900

2 0.60 10f

;

(to give 350Hz)f [4]

(c) (i) transfer of energy by means of vibrations/oscillations;

vibrations all in one direction parallel to direction of energy transfer; [2]

(ii) 330

350 or use of c f ;

0.94m ; [2]

Award [2] for bald correct answer.


Part 2 Diffraction of light

(a) (i) spreading out of light;

beyond that predicted by the geometric pattern / by the obstacle shape / OWTTE; [2]

(ii) diagram:

central symmetrical maximum;

at least one secondary maximum on each side with smaller height no more

than one third height of central maximum;

minima drawn to zero; (i.e. sitting on x-axis) [3]

(iii) 9

3

620 10

0.4 10b

;

9

3

2.0 1.9 620 10(2 )

0.4 10w D

;

5.9mmw ; [3]

Award [3] for bald correct answer.

(b) (i) the images can be seen separately; [1]

(ii) diffraction occurs (at the aperture/iris of the eye);

each lamp gives rise to a diffraction pattern (at the back of the eye/on the

retina);

(for distant lamps) the two diffraction patterns overlap;

so that patterns cannot be distinguished / OWTTE; [4]


SECTION B

B1. Part 1 Wave motion

(a) (i) 1.5 mm; [1]

(ii) 8.0 cm; [1]

(iii) distance travelled in 0.20 s is 3.2 cm;

so speed is 2

13.2 100.16ms

0.20

; [2]

(iv) 2

0.162.0Hz

8.0 10f

; [1]

(b) travelling waves transfer energy (standing waves do not);

travelling waves have a constant amplitude (standing waves do not);

standing waves have points that always have zero displacement (travelling waves

do not);

the phase of a travelling wave constantly changes (but in standing waves points in

between consecutive nodes have constant phase); [2 max]

(c) (i) it is the speed of energy transfer/rate/speed at which wavefronts move forward; [1]

(ii) a standing wave is formed from the superposition of two travelling waves;

wave speed refers to the speed of the travelling waves; [2]

(d) (i) the oscillating string collides with the air molecules surrounding it;

creating a pressure/longitudinal wave; [2]

(ii) wavelength of wave on string is 2 0.80 1.6m ;

frequency is then 240

150Hz1.6

;

sound has the same frequency and so wavelength is 340

2.3m150

; [3]

Award [1 max] for those using a wavelength of 0.80 m obtaining a wavelength

of 1.1 m in air. Accept alternative derivations that use a ratio and do not

calculate the frequency explicitly.

– 13 – M08/4/PHYSI/HP2/ENG/TZ2/XX/M

B3. Part 1 Wave phenomena (a) (i) C shown where graph line cuts x-axis; [1] (ii) time period 0.30 ms; =

use of v fλ= and 1 fT

= or vTλ

= ;

; [3] 3380 0.30 10 0.11mλ −= × × = ECF if time period misread. (b) (i) superposition of two waves / OWTTE; of same frequency and amplitude travelling in opposite directions; [2] (ii) stationary/standing wave is set up in the tube; heaps form at the (displacement) nodes / powder pushed away from antinodes; [2] (iii) wavelength ; (2 9.3 )18.6cm= × = speed 1(1800 0.186 ) 330ms−= × = ; [2] ECF if value of wavelength wrong. (c) heaps further apart means longer wavelength;

hence speed increases (as temperature rises);⎧⎨⎩

Do not award if there is no reasoning orreasoning is fallacious or misleading. [2]

(d) (i) when two waves meet; resultant displacement found by summing individual displacements; to give maximum displacement / displacement greater than that of an individual

wave; [3] (ii) line in correct position, labelled C; [1] (iii) line in correct position, labelled D; [1]

(e) use of axD

λ = and ; 24.0 10 ma −= ×

24.0 10 1.2

1.5λ

−× ×= ;

; [3] 23.2 10 mλ −= × ECF if value of “a” wrong [2 max].


B2. Part 1 Waves (a) Transverse the particles (of the medium) vibrate at right angles; to the direction of energy transfer; Longitudinal the particles (of the medium) vibrate in the same direction as the direction of

energy transfer; [3] (b) (i) time period = 0.13 s;

1 Hz ;0.13

fT1⎛ ⎞= = = 7.7 (±0.3) ⎜ ⎟

⎝ ⎠ [2]

Award full marks for bald correct answer. (ii) 8 mm; [1]

(c) ;vf

λ =

157.7

;

1.95 cm 2.0 cmλ = ≈ [2] (d) start at (−1.2 →−2.0) on y-axis; sine curve of amplitude 8 mm; wavelength 2 cm; [3]

(e) use of 1 1

2 2

sinsin

vv

θθ

=

22 1

1

sin sin ;vv

θ θ=

sin 3020=

15to give 2 ;θ °= 42

angle °= 48 ; [3]


(f) (i) each slit acts as a point source of waves; waves from these sources interfere; because of the principle of superposition; where a trough of one wave meets a crest of another the resultant displacement

will be minimum / waves arrives with opposite phase/completely out of phase / so destructive interference occurs; [4]

(ii) ;dsDλ

=

9.4 ;2.0

18×=

= 85 / 90 cm; [3] Award full marks for bald correct answer. (g) (i) the phase difference between them is constant; [1] (ii) fringes of equal intensity / intensity reducing from centre fringe; and equal width; [2] [max 1] if fringes do not touch axis. B2. Part 2 Magnetic fields (a) effective current in each side of coil = 20I;

each wire needs to produce 1 ;4

B

0

π ;rBIµ

2=

π 0.6 7.0 10 ;π 10 80

−5

−7

2 × × ×=

4 × ×

A= 2.6 [4]

Accept correct substitution for I into2

IBr

µ0=π

to show that 10B T−5= 7.0×

(b) plane vertical; plane normal to Earth field; [2]

Documents

Combined IB HL Waves Qs 0809 MS