6.3 Further Mechanics and Thermal Physics - Simple Harmonic Systems 2 – Mark schemes

Q1.

(a)      (1)

Oscillations must be of small amplitude (1)2

(b)     (i)      f =  (1)

[or T = = 1.8(6) (s)]

 [or ] (1)

l = 0.85(9)m (1)(allow C.E. for values of f or T)

(ii)     amax{= (–)(2πf)2A} = (2π × 0.538)2 × 51 × 10–3 (1)

(= 0.583 ms–2)(allow C.E. for value of f from (i))

Fmax(= mamax) = 1.2 × 10–2 × 0.583 (1)

         = 7.0 × 10–3N (1)(6.99 × 10–3N)

[or Fmax(= mg sin θmax) where sin θmax=  (1)

= 1.2 × 10–2 × 9.81 ×  (1)

= 6.99 × 10–3N (1)]6

[8]

Q2.(a)     (i)      elastic potential energy and gravitational potential energy ✓

For elastic pe allow “pe due to tension”, or “strain energy” etc.

1

(ii)     elastic pe → kinetic energy → gravitational pe                                                         → kinetic energy → elastic pe ✓✓ [or pe→ke→pe→ke→pe is ✓ only] [or elastic pe → kinetic energy → gravitational pe is ✓ only]

If kinetic energy is not mentioned, no marks.Types of potential energy must be identified for full credit.

2

(b)     (i)      period = 0.80 s ✓ during one oscillation there are two energy transfer cycles (or elastic pe→ke→gravitational pe→ke→elastic pe in 1 cycle) or there are two potential energy maxima per complete oscillation ✓

Mark sequentially.2

(ii)     sinusoidal curve of period 0.80 s ✓ – cosine curve starting at t = 0 continuing to t = 1.2s ✓

For 1st mark allow ECF from T value given in (i).2

(c)    (i)      use of T = gives 0.80 = ✓

∴ = 22 (21.6) ✓    N m–1 ✓

 Unit mark is independent: insist on N m–1. Allow ECF from wrong T value from (i): use of 0.40s gives 86.4 (N m–1).

3

(ii)     maximum ke = ( ½ mvmax2) = 2.0 × 10–2 gives

vmax2 = ✓ (= 0.114 m2s–2)  and  vmax = 0.338 (m s–1) ✓

         vmax = 2πfA gives A = ✓

                               and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm

[or maximum ke = (½ mvmax2) = ½ m (2πfA)2 ✓

        ½ × 0.35 × 4π2 × 1.252 × A2 = 2.0 × 10–2 ✓

∴ A2 = ✓ ( = 1.85 × 10–3 )          and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm ] [or maximum ke = maximum pe = 2.0 × 10–2 (J)     maximum pe = ½ k A2 ✓     ∴ 2.0 × 10–2 = ½ × 21.6 × A2 ✓

from which A2 = ✓ ( = 1.85 × 10–3 )      and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm ]

First two schemes include recognition that f = 1 / T i.e. f = 1 / 0.80 = 1.25 (Hz).Allow ECF from wrong T value from (i) – 0.40sgives A = 2.15 × 10–2m but mark to max 3.

Allow ECF from wrong k value from (i) –86.4Nm–1 gives A = 2.15 × 10–2m but mark to max 3.

4[14]

Q3.(a)      (i)     Maximum displacement (of carriage/pendulum from rest

position)

B11

(ii)     6.0 (m)

B11

(iii)    Clear evidence of what constitutes period

C1

4.8–4.9 (s)

A12

(b)     (i)     Use of v = 2πfA

C1

7.07 (ms–1)

A12

(ii)     Use of a = 4π2f 2A

C1

11.1 (ms–2) ecf

A12

(iii)    Substitution into or rearrangement of T = 2π√l/g

C1

3.98 (m)

A12

(c)     Applied frequency = natural frequency

B1

Mention or clear description of resonance

B12

(d)     Resistive/frictional/damping/air resistance forces

C1

due to friction in named place (eg in bearings)/air resistance acting onnamed part (allow ride/gondola here)

A1

low friction/large mass or inertia /streamline/smooth surface etc.

B13

[15]

Q4.(a)     (i)      use of mg = kΔL or F = kΔL and F = mg

C1

extension = 5.9 m

A1

total length = 25.9 m (allow 20 + their extension)

B13

(ii)     20 + twice (5.9) amplitude + 2.6; 34.4 m;allow ecf from ai

B11

(b)     (i)      T = 2π√m/k and T = 1/f or f = 1/2π √k/m

B1

correct substitution: allow for calculationof T (4.85 s)

B1

0.21 or 0.206 (Hz)

B13

(ii)     substitutes data in vmax = 2πfA

C1

5.4 ms–1 (5.28 to 5.53)

A12

(iii)     two complete oscillations shown with positiveand negative velocities and acceptable shape

(condone more than 2)

B1

and two fromperiod of 5 s used in graph (allow ecf for T fromearlier part)

B1

start at 0 and positive velocity change at T = 0with positive and negative velocities shown

B1

max velocity shown decreasing

B13

(c)     (i)      it would have to raised

B1

rest extension would be greater/rider would benearer the ground if extension unchanged

B1

the rider has to move down a distance = to theamplitude (5.9 m) from the new rest position

or with same initial extension/energy stored in rope,the rider would reach a lower heightamplitude would be loweror due to the larger mass more energy (= mgh) isneeded to reach the same height

so initial extension would have to be increased

B13

(ii)     the rope would become slack at the top of theride so the rider would go into free flight/riderwould overshoot the highest point

B1

the rider would fall and, with negligible airresistance, the rope would again absorb the energyarriving back at the start point or rider is more likelyto fail to reach the ground after one oscillationdue to energy losses/air resistance

B1

the PE gained (at the top of the flight) can (at most)only be converted back to the elastic energy that wasstored in the rope at the start

(allow a statement to the effect that to hit the floorwould contravene conservation of energy or require anenergy input)

B13

[18]

Q5.(a)    acceleration is proportional to displacement (from equilibrium)  

Acceleration proportional to negative displacement is 1st mark only.

acceleration is in opposite direction to displacementor towards a fixed point / equilibrium

Don’t accept “restoring force” for accln.

position  2

(b)    (i)       3SF is an independent mark.

 When g = 9.81 is used, allow either 0.502 or 0.503 for 2nd and 3rd marks.

    ] Use of g = 9.8 gives 0.502 Hz: award only 1 of first 2 marks if quoted as 0.502, 0.503 0.50 or 0.5 Hz.

answer to 3SF  3

(ii)      Allow ECF from any incorrect f from (b)(i).

= 0.42 (0.419) (m s−2)  2

(c)     recognition of 20 oscillations of (shorter) pendulum

and / or 19 oscillations of (longer) pendulum    

Explanation: difference of 1 oscillation or phase change of 2

or Δt = 0.1 so n = 2 / 0.1 =20, or other acceptable point  

time to next in phase condition = 38 (s)  Allow “back in phase (for the first time)” as a valid explanation.

[ or (T = 1.90 s so) (n + 1) × 1.90 = n × 2.00  

gives n = 19 (oscillations of longer pendulum)    

minimum time between in phase condition = 19 × 2.00 = 38 (s)   ]3

[10]

Q6.(a)     (i)      correct period read from graph or use of f=1/T 0.84±0.01

C12.4 Hz gets C1

correct frequency 1.2 (1.18 − 1.25 to 3 sf)

A1

(ii)     correct shape (inverse)

B1

Crossover PE = KE

B1

(b)    (i)       

C1

48.7 (49) m

A1

(ii)     v = 120 000 / 3600 = 33(.3) m s−1

B1

Use of F = m v2/r (allow v in km h−1 )

B1

Total tension = 6337 + (280 × 9.81) = 9.083 × 103 NAllow their central force

B1

Divide by 4      2.27 × 103 NAllow their central force

B1

(iii)    mgh = ½ mv2

B1Condone: Use of v = 2πfA (max2)

9.8×44 = 0.5 v2      Allow 45 in substitution

B1Condone         22 m s−1

29.4 m s−1          (Use of 45 gives 29.7)

B1

106 km h−1 (their m s−1 correctly converted)Or compares with 33 m s−1

B1

(iv)    1/16th(0.625) % of KE left if correct

M1Allow 1/8 (0.125)or 1/32(0.313)

KE at start = 5.6 × 104 J or states energy ∝ speed2 so speed is ¼

M1Allow for correct subn E =½ 280 × 202 x factor from incorrect number of swings calculated correctly

Final speed calculated = 5 m s−1

A1Must be from correct working

[17]

Q7.(a)     2 smooth curves to show envelope of exponential decay waveform; lines to be

continuous from first to fifth points, maximum deviation from best-fit lines thorough each set of 5 points must not be greater than 1 mm ✔

1

equilibrium position marked on grid with horizontal line at A = 15.7 ± 0.1 cm ✔

1

(b)     evidence of valid working (using the line(s) and/or the equilibrium position) established in (a)(iii) to test for the exponential nature of the decay (working may be shown on the graph): do not penalise confusion between n and timeeitherevidence of relevant A values [2A ie A–(–A)] measured from graph (correct to nearest mm) or deduced from difference between tabulated values and equilibrium position of pointer) or 0/3 1 ✔at least two half life measurements (expect evidence of working) 2 ✔values obtained giving n½ = 6.3 ± 0.3 from either or both curves confirming exponential decay 3 ✔or1 ✔ as above; evaluates at least two ratios of successive amplitudes [or the fractional change in successive amplitudes], eg

2 ✔; ratios obtained giving consistent results to

± 5% confirming exponential decay 3 ✔

or1 ✔ as above; evaluates difference between natural logs of at least two successive amplitudes, eg ln(A0) – ln(A1) and ln(A1) – ln(A2) ✔ differences obtained giving results consistent to ± 10% confirming exponential decay 3 ✔

35

Q8.

(a)      (i)     speed at P, v (= ) =    

= 22(.1) (m s–1)    2

(ii)     use of F = k∆L gives     

= 11 (10.5) (m)  2

(b)      (i)      period T = 2 = 2π   (= 6.51 s)

time for one half oscillation = 3.3 (3.26) (s)  2

(ii)     frequency   (= 0.154 (Hz))

use of v = ±2πf when x = 10.5 m and v = 22.1 m s–1 gives 22.12

= 4π2 × > 0.1542 (A2 – 10.52)    

from which A = 25.1 (m)  

        [alternatively, using energy approach gives ½ mvP2 + mg∆L = ½ k(∆L)2

 

        ∴ (29 × 22.12) + (58 × 9.81 × ∆L) = 27 (∆L)2

        solution of this quadratic equation gives ∆L = 35.7 (m)  

        from which A = 25.2 (m)   ]3

(c)     bungee cord becomes slack  

student’s motion is under gravity (until she returns to P)  

has constant downwards acceleration or acceleration is not ∝ displacement  2

(d)     (i)      when student is at R or at bottom of oscillation  1

(ii)     at uppermost point or where it is attached to the railing  

because stress = F/A and force at this point includes weight of whole cord  

[accept alternative answers referring to mid-point of cord because cordwill show thinning there as it stretches or near knots at top or bottomof cord where A will be smaller with a reference to stress = F/A]

2[14]

Q9.(a)     (i)      ∆E = mg∆h

B1

= (16.8+1.2)9.8 × 0.5 or a mass × 9.8 × 0.5

B1

= 88.2 (J)

B13

(ii)     108 J or answer to (a) (i) + 20 J

B11

(iii)     108/0.40 allow ecf from (ii) (i.e. their (ii)/0.40)

C1

270 N {68/.4 = 170}

A12

(b)     gain in KE = loss in PE - work done

C1

= 88-20 = 68

C1

KE = ½ mv2

C1

v = 2.7(5) m s–1 no ecf

A14

(c)    

 

graph starts at origin and forms a full rounded peak

B1

exactly two cycles (4 peaks) shown but not arches

B1

height of peaks decreases and peaks approximately equallyspaced

B13

[13]

 

Q10.(a)     (i)      for one spring, change in force ΔF = kΔL = 30 × 60 × 10–3

= 1.8 (N)  

resultant force (= [F + ΔF] – [F – ΔF]) = 2ΔF  

(= 3.6 N)

alternative using answer from (b) (ii)

a = (2πf)2x = (2π × 1.38)2 × 60 × 10–3 = 4.51 (m s–2)  

resultant force = ma = 0.80 × 4.51 (= 3.6 N)2

(ii)     acceleration a  =  = 4.5(m s–2)  to the right  

alternative for first mark using answer from (b) (ii)

a = (2πf)2 x = (2π × 1.38)2 × 60 × 10–3 = 4.5 (m s–2)  2

(b)     (i)      acceleration is proportional to displacement

(from equilibrium position)  

acceleration is in opposite direction to displacement[or acceleration is towards a fixed point/equilibrium position]  

2

(ii)        = (1.38 Hz)

period T  = 0.73 (0.726)  [or 730]

s  [ms]3

(c)     (i)      (Hz)  1

(ii)     vmax (= 2π fA) = 2π × 1013 × 10–11 = 630 (628) (m s–1)  1

(iii)     max EK (= ½ mvmax2) = ½ × 1.0 × 10–25 × 6282 = 2.0 × 10–20 (J)  

[or using ½ kA2 approach]1

[12]

Q11.(a)     (i)      toward B

B11

(ii)     15 × 0.20 = 3 mm

B11

(b)     (i)      period = 0.8 s

C1

use of T = 2π√L/g

C1

0.16 (0.159) m

A13

(ii)     lower initial displacement

B1

lower inertia/more likely to begin moving asthe Earth moves

B1

no effect

B1

period of a simple pendulum is independent of themass of the bob/mass of bob is not in the formulafor the period of a simple pendulum/period onlydepends on length (and g)

B14

(c)     (i)      clearly states consistency of ratios of successiveamplitudes as thetest

B1

one ratio of successive amplitudes correctlydetermined

B1

         two ratios correctly determined and conclusion

B13

(ii)     the oscillations are damped/air resistancementioned/friction of pen against paper

B1

energy is lost because of air resistance/work isdone against air resistance/energy lost movingair out of the way/giving air kinetic energy

B12

(iii)     it will come to rest quicker

M1

the bob loses a greater proportion of its energyduring each oscillation

A12

or pendulum has lower inertia so damping force hasgreater effect

or oscillating pendulum (initially) has less energy

or air resistance (initially) is unchanged[16]

Q12.(a)     acceleration is proportional to displacement (1)

acceleration is in opposite direction to displacement, or            towards a fixed point, or towards the centre of oscillation (1)

2

(b)     (i)      f =  = 1.1 Hz (or s–1) (1)               (1.09 Hz)

(ii)     (use of a = (2πf)2A gives)a = (2π × 1.09)2 × 76 × 10–3 (1)   = 3.6 m s–2 (1)           (3.56 m s–2)(use of f = 1.1 Hz gives a = 3.63 m s–2)(allow C.E. for incorrect value of f from (i))

(iii)     (use of x = A cos(2πft) gives)x = 76 × 10–3 cos(2π × 1.09 × 0.60) (1)   = (–)4.3(1) × 10–2m (1)        (43 mm)(use of f = 1.1 Hz givesx = (–)4.0(7) × 10–2 m      (41 mm))direction: above equilibrium position or upwards (1)

6

(c)     (i)      graph to show:          correct shape, i.e. cos curve (1)          correct phase i.e. –(cos) (1)

(ii)     graph to show:          two cycles per oscillation (1)          correct shape (even if phase is wrong) (1)          correct starting point (i.e. full amplitude) (1)

max 4[12]

Q13.(a)     (i)      (vertically) downwards [or top to bottom, or down the page] (1)

1

(ii)     force on sphere F (= kx) = 0.24 × 18 × 10−3 (1) (= 4.32 × 10−3 N)1

(iii)     use of F = EQ gives E =  (1) (= 1.05 × 105 V m−1)

use of E =  gives separation d =  (1)

= 4.8 × 10−2 (m) (1) (4.76 × 10−2)3

(b)     (i)      electric field becomes zero (or ceases to exist) (1)

flow of charge (or electrons) from one plate to the other[or plates discharge] (1)

(until) pd across plates becomes zero [or no pd across plates,or plates at same potential] (1)

max 2

(ii)     net downward force on sphere (when E becomes zero)[or gravitational force acts on sphere, or force is weight] (1)

this force extends spring (1)

force (or acceleration) is proportional to (change in)extension of spring (1)acceleration is in opposite direction to displacement(or towards equilibrium) (1)

for shm, acceleration  (−) displacement[or for shm, force  (−) displacement] (1)

max 3[10]

Q14.(a)     (grav) potential energy → kinetic energy → (grav) potential energy

→ kinetic energy → gravitational potential energy (1)

energy lost to surroundings in overcoming air resistance (1)2

(b)     (i)      period T =  = 2.8 s (1)

use of T = 2π  gives length l =  (1)

giving distance from pt of support to c of m, l = 1.9 (m)or 1.95 (m) (1)

answer must be to 2 or 3 sf only (1)4

(ii)     Ek = mgΔh stated or used (1)

gives Ek of girl at lowest point = 18 × 9.81 × 0.25 = 44 (J) (1)2

(iii)     ½ mv2 = 44.1 gives max speed of girl v =  = 2.2 (m s–1) (1)

[alternatively: A2 = (3.9 – 0.25) × 0.25 gives A = 0.955 (m)and vmax = 2π f A = (2π/2.8) × 0.955 = 2.1 (m s–1) (1)]

1

(c)     graph drawn on Figure 2 which:

shows Ek = 0 at t = 0, T/2 and T (1)

has 2 maxima of similar size (some attenuation allowed)at T/4 and 3T/4 (1)

is of the correct general shape (1)3

[12]

Q15.

(a)     (i)      

(ii) 

 

a correct (= 20mm) (1)

x = ±20 mm at t = 0 (1)

T correct (= 0.70 s)(1)(5)

(b)     (i)      vibrates at 0.5 Hz with low amplitude (1)

(ii)     vibrates with high amplitude (1)at natural frequency (1)resonates (1)

(max 3)[8]

Q16.(a)     (i)      mg = ke (1)

 = 61(.3) N m−1 (1)

(1) (= 0.667 s)

(ii)     (1)(= 1.50 Hz)4

(b)     (i)      forced vibrations (at 0.2 Hz) (1)amplitude less than resonance (≈ 30 mm) (1)(almost) in phase with driver (1)

(ii)     resonance [or oscillates at 1.5 Hz] (1)amplitude very large (> 30 mm) (1)oscillations may appear violent (1)

phase difference is 90º (1)

(iii)     forced vibrations (at 10 Hz) (1)small amplitude (1)out of phase with driver [or phase lag of(almost) π on driver] (1)

Max 6[10]

Q17.

(a)     use of mg = ke gives k = (1)

                                     = 56 N m–1 (1) [or kg s–2]2

(b)     (i)      28 (N m–1) (1) (unit to be given in either (a) or (b))              (allow C.E. from (a))

(ii)     (use of T = 2π gives) T = 2π = 0.84 (s) (1)

             (allow C.E. for value of k from (b)(i))

number of oscillations per minute = = 71 (1)

             (allow C.E. from (b)(ii))3

[5]

Q18.

(a)     (use of T = 2π  gives) T = 2π  (1)

= 1.8 s (1)2

(b)     mgh = ½ mv2 (1)

v =  (1) (= 0.63 m s-1)

vmax = 2πfA =  (1)

A =  (1) (= 0.18m)

[or by Pythagoras A2 + 7802 = 8002

gives A =  (1) ( = 180 mm)

(or equivalent solution by trigonometry (1) (1))

vmax = 2πfA or =  (1)

=  (1) (= 0.63 m s-1)4

(c)     tension given by F, where F – mg =  (1)

F = 25 × 10-3  = 0.26 N (1)2

[8]

Q19.(a)     accelerating to left (1)

net force to left on M due to compression on right and tension on left (1)/p>(2)

(b)     slider is at 33/50 of length (1) (uniform track so) resistance ∝ length (1)

V = × 5.0 (V) (1) = 3.3 V (1)

ΔV = +0.8V (1)(max 4)

(c)     damping (1) to prevent (reduce) oscillation of mass (during changes in motion) (1)

(2)[8]

Q20.(a)     2.2 s

c.a.o.

B1                1

(b)     exactly two reasonable sine wave cycles drawn

B1

displacement = 10 cm when time = 0

B1

time = 2.2 s after one cycle

B14

peaks decrease to approximately 7.0 cm after two cyclesor 8.4 cm after one cycle

B1

award two marks if half-cycle confused with full cyclebut otherwise correct

(c)     (i)      the period would be decreased

B1

(ii)     there would be less damping/more oscillations beforethe pendulum comes to rest

B12

[7]

Q21.(a)     (i)      period = 1.8 or 1.9 s or f = 1/T

C1

0.56 (0.556)Hz or 0.53 (0.526) s if T = 1.9 s

A1

(ii)     0.074 – 0.078 m

B15

(iii)     frequency remains constant

B1

amplitude reduces

B1

(b)     attempt shows understanding of π/2 phase difference

(lag or lead)

M1

constant phase difference and amplitude

(acceptable quality)

A12

(c)     (i)      maximum acceleration = ω2A or ω = 2πf

C1

0.91(3) m s−2 or 0.83 if T = 1.9 s

(ecf from (a) (i)) and (a) (ii))

(not allowed if period given as answer in (i))

A1

(ii)     (maximum) speed = ωA (0.267 m s−1)

C1

use of KE = ½ mv2 with at least m (= 8 × 10−3)substituted

C1

(2.5 to 3.0) × 10−4 J

A15

or

maximum restoring force = 8.0 × 10−3 × 0.91

C1

oscillator energy = ½ × F × A or0.5 × 8.0 × 10−3 × 0.91 × 0.075

C1

(2.5 to 3.0) × 10−4 J

rA1[12]

Q22.(a)     ½ Fx or ½ kx2

C1

29.4 mJ

A12

(b)     (i)      amplitude clearly marked on diagram - must touchlines or be an accurately drawn equivalent distance

B11

(ii)     idea of interchange of p.e. and k.e.

B1

appropriate use of elastic p.e. at start of cycle and ofgravitational p.e. highest point + some k.e. in between.

B1

2[5]

Q23.(a)     a = v2/r

C1

          = (7.68 × 103)2/(6.760 × 106) or r = 6380 + 380

C1

          = 8.73 m s–2

A13

(b)     (the scientist is in) free-fall (owtte)

B1

          his/her weight provides the centripetal force

B1

(to maintain) the same orbit/same radius and velocity/sameacceleration (as the ISS)

B1

          his/her body experiences no motion/force relative to the ISS

B1max 2

(c)     (i)      k = 4π2m/T2 or T = 2π(m/k)1/2 and inferred transposition

B1

                                                               = 4 × (3.142)2 × 2.0/ (1.2)2

(= 54.8)

B12

(ii)     use of T = 2π(m/k)1/2 and f = 1/T

C1

f = 5.4 × 1012 Hz

A12

[9]

Q24.(a)     (i)      Unchanged

B11

(ii)     ½ OWTTE

B11

(iii)     T = 2π√ (M/k)

B1

T2 = 4π2 × M/k   (square and re-arrange)

B12

(iv)    T = 1/0.91 [= 1.1 s]

C1

1.12 × 190000 /4π2

A1

So mplatform = (cand ans for M –5300) leading tocorrectly evaluated answer

B13

(b)     v shape correct            [cos graph]

B2

or v shape inverted [-cos graph]

B1

          k.e. always +ve

M1

k.e. freq doubles

M1

k.e. shape acceptable

A15

(c)     max 4 from:mention of forced oscillationplatform frequency always matches lorry’s frequencymention of resonancesmall amplitude when well away from resonant frequencylarge amplitude at resonance          [do not infer small amppoint from this] resonant freq close to 0.91 Hz

B44

[16]

Q25.(a)     2π/T or 2πf or angular speed/velocity/frequency/Δθ ÷ Δt with

symbols defined

B1

displacement direction opposite to acceleration vector/acceleration towards central point/equilibrium point

B12

(b)     (i)      ω = 2π/T = 2.86 rad/s can appear as (2π/2.2) in

C1

substF = 0.053(1) N

A12

(ii)     to centre of turntable/rotation/circle not ‘towardscentre’

B11

(c)     (i)      l = [T2g/4π2] = 1.20 m

A11

(ii)     correct use of a = ω2A

M1

or accel = v2/r or F/m approacha = 1.0 / 1.1 / 1.04 / 1.06 m s–2 [cao]

A12

(d)     a origin at zero

C1

a in antiphase

A1

k.e always positive and start at maximum

C1

k.e. twice f and good shape

A14

[12]

Q26.(a)     (i)      acceleration (not a) and displacement (not x) are in

opposite directions OR restoring force/accelerationalways acts toward rest position

B11

(ii)     (+) sine curve consistent with a graph

B11

(b)     (i)      statement that EK = EP

B1

statement of max values considered

B1

EP = ½ k(Δl)2 or EPmax = ½ kA2

B1

correctly substituted values

B1

EK = 3.7 × 10–2 J

B1

ORf = 1/T or T = 3.97 s or period equation

B1

leading to f = 0.252 Hz

B1

ωmax = 1.58 rad s–1 or vmax = 0.055ms–1 (seen or used)

B1

substituted values into  EK = ½mA2ω2 or EK = ½mv2

B1

EK = 3.7 × 10–2 J

B15

(ii)     any attenuation from t = 0 seen

M1

10 mJ or E0/4 at either 4s or third hump

M1

consistent period values minima at 1 and 3smaxima at 0 and 4s

A13

[10]

Q27.(a)     displacement negative cosine

B1

velocity consistent with first graphB1

acceleration consistent with first or second graphB1

at least one cycle, constant amplitude (condone small decay ), include A for displacement, reasonably drafted

B1

(b)     use of T = 2π i.e. substituted values or 0.74 seenC1

use or implied use of T = C1

1.34 HzA1

[7]

Q28.(a)     Time for one cycle

M1

One cycle defined correctly in terms of diagram, canbe on diagram

A12

(b)     B

B1

Mention of air resistance, allow drag OR bob faster incentre of motion

B1

Links two ideas

B13

[5]

 

Q29.(a)     force is needed toward the centre or there is acceleration

toward the centre

B1

movement to the left/toward A/away from the centre(or indicated on diagram)

M1

right hand spring (attached to B) has to stretch toprovide force

A13

(b)     (i)      acceleration = v2/r or speed = 12.5 m s–1

or a = r ω2 and v = rω or ω = 0.52 rad s–1 or 452/0.024

C1

6.5 m s–2 8.4 × 104 km h–2 unit essential

A12

(ii)     Force on mass = 0.35 × (i) (2.28 N if correct)or use of F = mrω2 (0.35 × 24 × 0.522)

C1

0.82mm or 0.83 mm if (i) is correct;Movement = 12.6 × (i) mm

A12

(c)     (i)      T = 2π  or a = (2πf)2A or f = 1.4Hz or ω = 8.9 rad s–1

C1

k = 27.8 N m–1 use of T = 1/f or 2π/ω

C1

0.71 s (allow 0.70 s to 0.72 s)

A13

(ii)     sketch showing amplitude reducing with timestarting at max ignore changing period

B1

labelled consistently with answers to (b)(ii) and (c)(i).(0.71 s and initial displacement 82 mm)condone only one period shown correctly

B12

[12]

 

Q30.(a)     period time for one complete oscillation

B1

amplitude maximum displacement from undisturbed (equilibrium) position

B1(2)

(b)     clear description + repeat at least 10 times overall + averaging processB1

beginning or end measurement at equilibrium / use of datalogger (explicit) / fiducial mark

B1(2)

(c)     same time periodB1

significant and gradual reduction in amplitude compared to originalB1

(2)[6]

Q31.(a)     (i)      period = 1.2 s        or T = l / f

C1

0.83 Hz or 0.833A1

(2)

(ii)     period / frequency is the same or T1 = T2

B1

since T depends on – or T = 2π and k is constant(2)

(iii)    waveform sinusoidal or fits x = A sin ωt (accept cos waveform)M1

A after 1 / 8 cycle should be A /

or        find ω (= 2πf) and A; calculate x at any time and compareA1

or        use gradients to plot v–t then a–t graphsM1

          check whether for all t is constantA1

(2)

(b)     (i)      correct curvature with values at ends and centre correct and crossing at 0.08 ± 1 small square both sides

B1(1)

(ii)     correct curvatureM0

end displacements correct (1 / 2 that for X i.e. 0.1)A1

maximum correct (1 / 4 that for X i.e. 0.04 ± 1 small square)A1

(2)

(c)     reads maximum energy and displacement correctly 0.16 J and 0.20 m (allow e.c.f. for (b)(i) and (b)(ii))

C1

quotes E = k∆χ2 or E = Fx and F= kxC1

correct substitution leading to a value for k (8.0 N m–1)A1

or              maximum speed of oscillator X = ωA                  (using maximum gradient ≈ 0.2 / 0.15 or 27πfA (1.04 m s–1))

C1

                  mass of oscillator from Em = ½mv2 gives 0.189 kg                  (v = 1.04 gives 0.296 kg)

C1

                  T = 2π leading to a value for k(k = 5.2 N m–1)

                  (v = 1.04 gives 8.1 N m–1)A1

(discrepancy due to difficulty of measuring gradient)(3)

[12]

Q32.(a)     (i)      system made to oscillate by a periodic external force / energy source

or system made to oscillate by another oscillator it is connected to or another oscillator forcing the system to vibrate or an oscillator forcing another oscillator to vibrate

due to its own vibrationsA1

(1)

not made to oscillate by an external force not when its vibrations are not at its natural frequency not when it is made to oscillate by another system

(ii)     frequency of wheel (motor)= natural frequency of system condone ‘fundamental or resonant frequency‘ of the system or rotating wheel rotates at the resonant / fundamental frequency of the system

must be reference to the wheel or motor or driver oscillator not just when it vibrates at its natural frequency

B1(3)

(b)     T = 1.25 sB1

T = 2π(m / k)1 / 2 or numerical substitutionC1

0.36 kg allow e.c.f. for T; 0.6 s gives 0.82 kg

A1(1)

(c)     (i)      correct process using ratios (eg times to halve marked on the graph)

B1

three amplitudes read correctly or two times to halve read correctly

B1

ratios determined and conclusion drawn that it is exponential or a clear statement that the time to halve is the same for points indicated and conclusion that it is exponential

B1(3)

(ii)     amplitude correct for 25 s , i.e. 4 to 4.2 cm or energy proportional to A2 (allow 0.5 kA2 or 0.5 kx2)

C1

allow e.c.f. from (b); T = 0.6 s leads to amplitude = 6.8 cm ; ratio = 0.46

0.16-0.19 or 17/100 (2 s.f. only)A1

(3)[10]

Q33.(a)     force / acceleration proportional to displacement /

distance from mean positionB1

directed towards mean / fixed positionB1

(2)

(b)     (i)      0.96 s to 0.98 sB1

(1)

(ii)     1.02 Hz to 1.04 Hz                   e.c.f. 1 / (i)B1

(1)

(iii)    T = 2π√m / k or m = T2k / 4π2 in symbols or numbers, seen or usedC1

98 kg to 102 kg                        e.c.f. 106 × (i)2

A1(2)

(iv)    a = (–)ω2 AC1

1.03 m s–2 to 1.07 m s–2          e.c.f. 0.99(ii)2

A1(2)

(c)     time period (of oscillation caused by road markings) = s / v or 1.2 / 7 or 0.17 sC1

frequency = 1 / T or 5.8 Hz(use of v = fλ loses both of the 1st two marks)

A1applied frequency / time period is different from natural / resonant frequency so no resonance

B1(3)

(d)     (i)      KE at PB1

PE at QB1

at R, (nearly all of) energy absorbed by shock absorber / dissipated as internal energy (condone heat) in shock absorber / surroundings (allow lost in damping)

B1(3)

(ii)     energy proportional to (amplitude)2

C1at P, A = 2.5;at t, A = 1.0 or 0.9

C15.3 / 6.3 = 0.84 of energy absorbed at t (or 0.90, consistent with value of

A at t)A1

(3)[17]

Q34.(a)     (i)      free oscillation:

there is no force acting other than internal forces

or only internal forces are involved or there are no external forces acting

or no energy input (following initial displacement)

do not allow a free oscillation is not forced to oscillateB1

forced oscillation:

oscillator is acted on by a periodic external force(condone acted on by external varying force)

or energy is given periodically by an external source

or made to oscillate at the frequency of another oscillatorB1

(2)

(ii)     T = 2π√(l / g) or 8 = 2π√(l / 9.8)C1

15.9 (16) mA1

(2)

(iii)    a(max) = ω2 A (not x, s or r)C1

ω = 2πf or 2π / T or π / 4 or 0.785 (0.79) (radian s–1)C1

3.08 – 3.12 m s–2

A1(3)

(iv)    F = ma or F = 1.5 × 103 × (iii)C1

4 600 N to 4700 N ( but allow e.c.f. for a from (iii))A1

(2)

(b)     graph showing two cycles of energy change in a time of 8 s;peak at 12 kJ; shape always positive

M1

approx. sin2 shape; clearly curved at each zero value

(condone any extension beyond 8 s)A1

(2)[11]

Q35.(a)     acceleration/force is directed toward

a (fixed) point/the centre/the equilibrium positionora = –kx + ‘–’ means that a is opposite direction to x

B1

acceleration/force is proportional to the distance from thepoint/displacement

ora = –kx where a = acceleration; x = displacement andk is constant

B12

(b)     (i)      3.2 = 2π√l/9.8 (condone use of g = 10 m s2 for C mark)(use of a = –ω2x is a PE so no marks)

C1

2.5(4) m

A12

(ii)     Correct value at 0.5 m and correct curvature

M1

Energy at 1 m = 160 J

A12

[6]