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Sight and Waves Part 1

Mr. Klapholz Shaker Heights High School. Sight and Waves Part 1. Problem Solving. Problem 1. If a pipe is 85 cm long, and open on both ends, what is the lowest frequency standing wave that it can make? Take the speed of sound in air to be 340 m s -1 . Solution 1. Solution 1.

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Sight and Waves Part 1

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  1. Mr. Klapholz Shaker Heights High School Sight and Waves Part 1 Problem Solving

  2. Problem 1 If a pipe is 85 cm long, and open on both ends, what is the lowest frequency standing wave that it can make? Take the speed of sound in air to be 340 m s-1.

  3. Solution 1

  4. Solution 1 For the fundamental, half of a wavelength fits in the pipe. L = l/2. l = 2L = 2(0.85 m) = 1.7 m F = v / l F = 340 ms-1 / 1.7m F = 200 Hz

  5. Problem 2 The speed (v) of a wave traveling on a string is given by the expression: v = { T / m }½ • Deduce and expression for the frequency of the fundamental in a string of length L. • Use your answer to part (a) to estimate the tension in the ‘A’ string of a violin (frequency = 440 Hz).

  6. Solution 2, part (a) Imagine the fundamental mode on a string. How does the wavelength compare to the length of the string? • = 2L. f = v / l = {T/m}½ / 2L

  7. Solution 2, part (b) We already know that f = v / l = {T/m}½ / 2L. T = 4L2mf2 Now we need to estimate the length and the mass density of the string. L ≈ 0.5 m m ≈ 2 grams per meter = 2x10-3 kg m-1 T = 4L2mf2 = 4(0.5)2(2x10-3)(440)2 T = 400 N

  8. Problem 3 A person in a concert hall is listening to a violin play the note ‘A’ (440 Hz). The violinist is moving toward the listener at 30 m s-1 ! The listener perceives the sound as 484 Hz. What is the speed of sound in the air?

  9. Solution 3 f’ = f × { v / (v ± us) } • = 440 × { v / (v - 30) } 484 (v - 30) = 440v 484v –(484)(30) = 440v 484v – 440v = (484)(30) 44v = 14520 v = 330 m s-1

  10. Problem 4 A star is moving away from earth at 3.0 x 105m s-1. The light emitted from the star has a frequency of 6.0 x 1014 Hz. By how much will the frequency change?

  11. Solution 4 Df’ = f × ( v / c ) Df’ = 6.0×1014 × ( 3.0×105 / 3.0×108 ) Df’ = 6.0×1014 × ( 1.0×10-3 ) Df’ = 6.0×1011 Hz This is very small compared to the original frequency.

  12. Problem 4 If you shine Red light through a yellow filter, what comes out (if anything)?

  13. Solution 4 Red. (The only colors that pass through a Yellow filter are Red and Green.)

  14. Tonight’s HW: Go through the Sight and Waves section in your textbook and scrutinize the “Example Questions” and solutions. Bring in your questions to tomorrow’s class.

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