1. Standing Waves Resonance

2. Standing waves in Strings An incident wave undergoes fixed end reflection Standing waves produce nodes where the amplitude is zero Interference produces a wave that appears stationary with nodes and antinodes

3. Standing waves in Strings A symmetrical patern is stationary if: –Frequency doesn’t change –Distance between the two ends is constant If the frequency increases, the wavelength decreases so nodes get closer together. At certain frequencies (determined by the length of the medium) standing waves can set up to create antinodes (max amplitude) and nodes. Earthquakes in buildings achieve maximum destruction if a standing wave exists in the building!!

4. Standing waves in Strings The 1 st harmonic satisfies L = ½ λ with L as the string length At the fixed ends, there are always nodes The number of antinodes matches the harmonic number ƒo

5. Standing waves in Strings The 2 nd harmonic satisfies L = λ There are 3 nodes and 2 antinodes This is also called the 1 st overtone (1 above the fundamental frequency) 2ƒo

6. Standing waves in Strings The 3 rd harmonic satisfies L = (3/2) λ There are 4 nodes and 3 antinodes This is also called the 2 nd overtone (2 above the fundamental frequency) 3ƒo

7. Standing waves in Strings 1 st harmonic Fundamental frequency ( ƒ o ) 2 nd harmonic 1 st overtone 2ƒo 3rd harmonic 2 nd overtone 3ƒo

8. Standing waves in Strings Nodes are always ½ λ apart The general formula can be derived: n = 1, 2, 3…the harmonic number This same relationship applies to open end air columns (like with woodwind instruments)

9. Closed End Air Columns A tube filled with water can be slowly drained out while holding a tuning fork vibrating over th open end. At certain air column lengths, standing waves will be set up (with an anti-node at the top) and a loud sound is heard at this resonance point. Fixed end reflection of the sound occurs from the water surface Resonances occur at multiples of the wavelength at specific lengths.

10. Closed End Air Columns 1 st harmonic L = ¼λ ƒo 2 nd harmonic L = ¾λ 2ƒo 3rd harmonic L = 1¼ λ 3ƒo

11. Open End Air Columns Reflection with open tubes occurs due to pressure differences (air below tube is free to move) allowing open end reflection. Standing waves occur with the same geometry as waves in strings. Note: we use transverse waves to MODEL how sound behaves but really the patternd are more complicated to draw with longitudinal waves!!

12. Open End Air Columns 1 st harmonic ƒo 2 nd harmonic 1 st overtone 2ƒo 3 rd harmonic 2 nd overtone 3ƒo

13. Closed End Air Columns General Formula: n = harmonic number: 1, 2, 3. etc

14. Questions Ex 1) The first resonant length of a closed air column occurs when the length is 18 cm. a)What is the wavelength of the sound? b)What is the speed of sound if a 512 Hz source is used? a) First resonance occurs: L = ¼ λ = 18 cm. λ = 72 cm b) v = f λ = 512 Hz (0.72 m) = 3.7 x 10 2 m/s

15. Questions Ex 1) A tuning fork is held over a column filled with water. When the water is drained, the first resonance is heard at an air column length of 9.0 cm. a)What is the wavelength of the tuning fork? b)What is the air column length for the 2 nd resonance? c)Find the frequency of the tuning fork if it’s 20.0 o C.

16. Questions a) First resonance occurs: L = ¼ λ = 9.0 cm. λ = 36 cm b) L 2 = ¾ λ = ¾ (36 cm) = 27 cm c) v=f λ 332 m/s + 0.6 m/s/ o C(T) = f(0.36 m) F = 9.6 x 10 2 Hz

17. Questions A 500. Hz tuning forl is used to determine the resonances in an adjustable air column of air at both ends. The length of the air column changes by 33.4 cm between resonant sounds. a)Find the wavelength of the sound wave. b)Find the speed of sound. c)Determine the air temperature. Answers: 66.8 cm, 334 m/s, 3.33 o C

18. Questions Solution: a)Resonances occur ½ λ apart. 1/2 λ = 33.4 cm λ = 66.8 cm b)v = f λ v = 500 Hz (0.668 m) = 334 m/s c)v = 332 m/s + 0.6 m/s/ o C (T) 334 m/s = 332 m/s + 0.6 m/s/ o C (T) T = 3.33 o C