13548 10 The speed of sound was measured using a resonance tube arrangement as shown in Fig. 10.1. A loudspeaker was placed above the open end of the tube. The length L of the tube is 75 cm. The signal frequency was raised from 0 Hz until an exceptionally loud sound was heard for the first time. This frequency is called the fundamental frequency fo. loudspeaker resonance tube to signal generator L 75 cm Fig. 10.1 (a) Resonance is the effect which results in an exceptionally loud sound being heard. Explain why resonance occurs. [2] 13548 (b) The method described will identify the first position of resonance. (i) On Fig. 10.1 label the position of a node with the letter N and the position of an antinode with the letter A for the first position of resonance. [1] (ii) Describe how the air particles in the resonance tube are behaving at N and A. N A [2] (iii) The fundamental frequency fo is 114 Hz when the first position of resonance is identified. Use this information to calculate the speed of sound in air. Speed =                           m s-1 [3] (iv) The frequency is increased from 114 Hz until the next position of resonance is identified. Calculate the frequency at which this will occur. Frequency =               Hz [2] THIS IS THE END OF THE QUESTION PAPER 13548 BLANK PAGE DO NOT WRITE ON THIS PAGE 13548 BLANK PAGE DO NOT WRITE ON THIS PAGE Permission to reproduce all copyright material has been applied for. In some cases, efforts to contact copyright holders may have been unsuccessful and CCEA will be happy to rectify any omissions of acknowledgement in future if notified. DO NOT WRITE ON THIS PAGE SPH21/5 276070 For Examiner’s use only Question Number Marks 1 2 3 4 5 6 7 8 9 10 Total Marks Examiner Number DATA AND FORMULAE SHEET Physics [SPH11/SPH21] ADVANCED SUBSIDIARY General Certifi cate of Education Assessment Units AS 1 and AS 2 11378.02 11378.02 2 Data and Formulae Sheet for AS 1 and AS 2 Values of constants speed of light in a vacuum c = 3.00 × 108 m s–1 elementary charge e = 1.60 × 10–19 C the Planck constant h = 6.63 × 10–34 J s mass of electron me = 9.11 × 10–31 kg mass of proton mp = 1.67 × 10–27 kg acceleration of free fall on the Earth’s surface g = 9.81 m s–2 electron volt 1 eV = 1.60 × 10–19 J the Hubble constant H0 ≈ 2.4 × 10–18 s–1 Useful formulae The following equations may be useful in answering some of the questions in the examination: Mechanics conservation of energy 1 2 mv 2 – 1 2 mu 2 = Fs for a constant force Waves two-source interference λ = ay d diffraction grating d sinθ = nλ 11378.02 3 Light lens equation 1 u 1 v 1 f + = Electricity terminal potential difference V = E – Ir (e.m.f., E; Internal Resistance, r) potential divider Vout = R1Vin R1 + R2 Particles and photons Einstein’s equation 1 2 mv max 2 = hf – hf0 de Broglie equation λ = h p Astronomy red shift z = Δλ λ recession speed z = v c Hubble’s law v = H0 d 11378.02 4