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Physics 101 LO 6

Physics 101 LO 6

- 1. STANDING WAVES ON VIOLIN STRINGS
- 2. STANDING WAVES • Are stationary (as opposed to travelling waves) • Vs STANDING WAVE TRAVELLING WAVE
- 3. STANDING WAVES • Are the superposition of two harmonic waves with equal amplitude, frequency and wavelengths but moving in opposite direction v v Resulting Standing Wave from adding the two harmonic waves
- 4. STANDING WAVES • Can be generated by plucking a string with both ends fixed • Nodes are points with zero amplitudes • Antinodes are points with maximum amplitudes
- 5. STANDING WAVES ON STRINGS • Strings with two fixed ends can only produce standing waves with an integral number of half wavelength called normal modes • λ 𝑛 = 2𝐿 𝑛 where L = string length n = number of antinodes = 1, 2, 3, 4, … • The fundamental frequency (1st harmonic) is the lowest frequency (longest wavelength) • 𝑓1 = 1 2𝐿 𝑇 𝜇 where T = tension in the string 𝜇 = linear mass density of the string = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 • The allowed frequencies are called harmonics • 𝑓𝑛 = n𝑓1 n = 1, 2, 3, 4, …
- 6. QUESTION PART 1 Tom wants to make a violin for his sister as a birthday present. Violins usually make sound frequencies ranging from 200~3000Hz. He has a few 30 cm long strings with linear mass densities: A 2.8 × 10−4 kg/m B 4.0 × 10−4 kg/m C 0.62 g/m Which string should he use to make the violin in order to get a fundamental frequency of 700Hz if the tension in the string is kept at 70 N?
- 7. Hints • What variables are given in the question? • The fundamental frequency (𝑓1), tension (T), and string length (L)
- 8. Hints • What variables are given in the question? • The fundamental frequency (𝑓1), tension (T), and string length (L) • Which equation to use when solving for linear mass density? • 𝑓1 = 1 2𝐿 𝑇 𝜇 where T = tension in the string 𝜇 = linear mass density of the string = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔
- 9. Solution — Tom should use string B 𝑓1 = 700 Hz T = 70 N L = 30 cm = 0.30 m 𝑓1 = 1 2𝐿 𝑇 𝜇 Solve for 𝜇 𝑓1 × 2𝐿 = 𝑇 𝜇 (𝑓1 × 2𝐿) 2 = 𝑇 𝜇 𝜇 = 𝑇 ( 𝑓1×2𝐿) 2 = 70 𝑁 (700 𝐻𝑧×2×0.30𝑚) 2 = 3.97× 10−4 kg/ m ≈ 4.0 × 10−4 kg/m
- 10. QUESTION PART 2 The violin string broke after a few weeks, but Tom doesn’t have anymore of the same string. If he uses a string with linear mass density of 4.7 × 10−4 kg/m, what should the tension be in the string in order to produce the same sound frequency (700 Hz)?
- 11. Hints • What variables are given in the question? • The fundamental frequency (𝑓1), linear mass density (𝜇), and string length (L)
- 12. Hints • What variables are given in the question? • The fundamental frequency (𝑓1), linear mass density (𝜇), and string length (L) • Which equation to use when solving for tension? • 𝑓1 = 1 2𝐿 𝑇 𝜇 where T = tension in the string 𝜇 = linear mass density of the string = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔
- 13. Solution 𝑓1 = 700 Hz 𝜇 = 4.7 × 10−4 kg/m L = 30 cm = 0.30 m 𝑓1 = 1 2𝐿 𝑇 𝜇 Solve for T 𝑓1 × 2𝐿 = 𝑇 𝜇 (𝑓1 × 2𝐿) 2 = 𝑇 𝜇 T = (𝑓1 × 2𝐿) 2 × 𝜇 = (700𝐻𝑧 × 2 × 0.30𝑚) 2 × 4.7 × 10−4 kg/m = 82.9 N ≈ 83 N
- 14. THANK YOU FOR WATCHING

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