Loudspeaker/driving frequency close or equal to its natural frequency

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Question nswer 1(a)(i) Resonance / resonating / resonates 1 (a)(ii) 1(b) Loudspeaker/driving frequency close or equal to its natural frequency so energy transfer is maximised/large energy transfer is very efficient Idea that energy would be transferred (from the glass) to the rubber band (as it deforms) work is done on the rubber band (by the glass) Some of the (transferred) energy becomes internal energy of rubber band some of the (transferred) energy is dissipated in the rubber band Total for Question 5 PhysicsndMathsTutor.com

Question (a) nswer Pendulum C has same/similar length as pendulum X Therefore C has the same/similar natural frequency as pendulum X idea that C is driven at its natural frequency (Hence) the energy transfer from X to C is most efficient There is a maximum transfer of energy from X to C correct reference to resonance (b) ny zero displacement point marked on original graph [do not insist on P ] (Minus) cosine graph drawn with same period as original graph 3 [Ignore amplitude of graph drawn] Examples of graphs: This candidate has identified P (although not used P ) and the cosine graph is well drawn. [ marks] This candidate has identified P correctly, and has drawn a minus cosine graph. Their graph starts from a time of T/4, which is just about acceptable. [ marks] This candidate has identified P correctly, but has drawn a sine curve. [1 mark] PhysicsndMathsTutor.com Total for question 5

Question 3(a) nswer 1 Use of F G m m r G = 6.6 10-11 (N m kg - ) [must see 6.6 10-11 when rounded to sf] 3(b)(i) 3(b)(ii) 1.5 10 7 N 0.3 G m 6.61 10 11 N m kg 160kg 0.75kg Read (peak) times from graph for at least 3 cycles T = 6.4 min (±0. min) [T = (380 ±1) s] [max 1 mark if correct answer shown without working] 8.0.5 min T 6.38min 4 ir resistance acts on the sphere [accept frictional forces (viscous) drag for air resistance] Energy is removed from the oscillation/system the oscillation/system is damped 3(b)(iii) [For mp do not credit energy is lost but accept energy is dissipated ; answer must indicate idea of transfer of energy] Evidence of values of at least 3 consecutive peaks read from graph [accept values of 3 points separated by equal time intervals] ttempt to obtain amplitudes, by subtracting 0.75 3 Calculation of two values of n+1 / n with corresponding conclusion Calculation of two values of difference of ln n+1 and ln n with corresponding conclusion Use peaks of graph to sketch curve Use curve to determine half-life [accept other ratio] Calculation of two values of half-life with corresponding conclusion 0 = 1.45-0.75 = 0.7, 1 = 0.75-0.5 = 0.5, = 1.1-0.75 = 0.35, 4 = 0.75 0.5 = 0. PhysicsndMathsTutor.com

1 0 1 3 0.50 0.71 0.70 0.35 0.50 0.70 0.5 0.71 0.35 Total for question 9 Question 4(a)(i) 4(a)(ii) nswer Resonance The vibrations from the engine/road surface/wheels must drive/force the tiger s head (to vibrate) at a frequency equal/close to its natural frequency 4(b)(i) Driver/forcing frequency Matches natural frequency Use of Use of a max = ω mplitude = 10 - m 3 4(b)(ii) Correct shape and phase (in antiphase with acceleration) for graph mplitude (ecf from (b)(i)) and a time marked on axes Total for question 8 PhysicsndMathsTutor.com

Question nswer *5 QWC Work must be clear and organised in a logical manner using technical wording where appropriate Max 5 Reference to resonance The sounding box is forced to vibrate (at the frequency of the tuning fork) Tuning fork and sounding box have similar natural frequencies Energy transferred from the tuning fork to the box The sounding box sets a large amount/mass/volume of air into vibration (Hence) the sound (wave) produced (in the air) has a larger amplitude Sounding box dampens the vibration (of the tuning fork) Larger rate of transfer of energy (to the air) means that the vibration persists for a shorter time 5 Total for question 5 Question 6(a) nswer k Use of ω = and T = π OR use of T = π m ω Time period = 0.43 s [allow any value that rounds to 0.4 s] m k 10 N m 1 1 k ω = = = 14.8rads m 0.55 kg π rad T = 0.45s 1 14.8rad s = 6(b) (i) 6(b) (ii) Energy of the system is dissipated or energy is removed from the system (by frictional forces) (Hence) the amplitude reduces Sinusoidal graph with at least cycles Decreasing amplitude pproximately constant time period 3 Total for question 7 PhysicsndMathsTutor.com

Question nswer 7 M 3 Curve : The system has a maximum amplitude at a particular frequency This is an example of resonance Resonance occurs when the forcing frequency is at (or near to) the natural frequency of the system t resonance there is an efficient/maximum transfer of energy (to the massspring system) MX 3 Curve B: B has a smaller amplitude than (for a wide range of frequencies) The modified system has (greater) damping Energy is being removed from the system The frequency at which resonance occurs is lower for the damped system 4 Total for question 4 PhysicsndMathsTutor.com

Question 8(a) nswer 1 Use of F G m m r G = 6.6 10-11 (N m kg - ) [must see 6.6 10-11 when rounded to sf] 8(b)(i) 8(b)(ii) 1.5 10 7 N 0.3 G m 6.61 10 11 N m kg 160kg 0.75kg Read (peak) times from graph for at least 3 cycles T = 6.4 min (±0. min) [T = (380 ±1) s] [max 1 mark if correct answer shown without working] 8.0.5 min T 6.38min 4 ir resistance acts on the sphere [accept frictional forces (viscous) drag for air resistance] Energy is removed from the oscillation/system the oscillation/system is damped 8(b)(iii) [For mp do not credit energy is lost but accept energy is dissipated ; answer must indicate idea of transfer of energy] Evidence of values of at least 3 consecutive peaks read from graph [accept values of 3 points separated by equal time intervals] ttempt to obtain amplitudes, by subtracting 0.75 3 Calculation of two values of n+1 / n with corresponding conclusion Calculation of two values of difference of ln n+1 and ln n with corresponding conclusion Use peaks of graph to sketch curve Use curve to determine half-life [accept other ratio] Calculation of two values of half-life with corresponding conclusion 0 = 1.45-0.75 = 0.7, 1 = 0.75-0.5 = 0.5, = 1.1-0.75 = 0.35, 4 = 0.75 0.5 = 0.5 PhysicsndMathsTutor.com

1 0 1 3 0.50 0.71 0.70 0.35 0.50 0.70 0.5 0.71 0.35 Total for question 9 Question nswer 9(a) Resonance System driven at / near its natural frequency () 9(b)(i) ny zero velocity point 9(b)(ii) ny maximum/minimum velocity point 9(c) Select 70 mm distance from passage/see 35 mm Use of a = -ω x Use of v = ω Correct answer (1 : (4) ω = 1 0.89 ms = 5.04rads 3.5 10 m 1 v = ω = 5.04s 1 3.5 10 m = 0.18 ms 1 9(d) QWC The answer must be clear and be organised in a logical sequence The springs/dampers absorb energy (from the bridge) (Because) the springs deform/oscillate with natural frequency of the bridge Hence there is an efficient/maximum transfer of energy Springs/dampers must not return energy to bridge / must dissipate the energy (max 3) Total for question (11) PhysicsndMathsTutor.com