1. Sound Test Answers

2. Question 1 What is the frequency of the pendulum given the graph of its horizontal position as a function of time? Show your work. Frequency = 2 cycle 6 seconds =.333 Hz

3. Question 2 What is the period of the pendulum given the graph of its horizontal position as a function of time? Show your work or explain your reasoning. Period = the time it takes to complete 1 cycle. It took 6 seconds for this pendulum to complete 2 cycle, 3 seconds for 1 cycle

4. Question 3 What is the length of this pendulum if the acceleration due to gravity is 9.8 m/s 2 ? T p = 2  L g

5. Question 3 What is the length of this pendulum if the acceleration due to gravity is 9.8 m/s 2 ? T p 2 = (2   L g

6. Question 3 What is the length of this pendulum if the acceleration due to gravity is 9.8 m/s 2 ? T p 2 g = L (2   (3.0s) 2 9.8 m/s 2 = L = 2.2 m (2  

7. Question 4 a) What would happen to the period and the frequency of oscillation if the acceleration due to gravity changed to 19.6 m/s2? Explain your reasoning.

8. Question 4 What would happen to the period and the frequency of oscillation if the acceleration due to gravity changed to 19.6 m/s 2 ? Explain your reasoning. T p = 2  L g

9. Question 4 Since the period (Tp) is inversely related to the square root of the acceleration due to gravity (g) the period will decrease with an increase in acceleration. The frequency will increase because T = 1/f T p = 2  L g

10. Question 4 b) What would happen to the frequency if the if the mass of the pendulum bob were quadrupled? The period of a pendulum is independent of the mass of the pendulum and therefore will remain the same.

11. Question 5 The amplitude of the mass on the string would be 1.0 m up to the peak Or The amplitude of the mass on the spring would be 1.0 m down to the trough

12. Question 6 A)What would happen to the period and frequency of oscillation for this mass spring system if the acceleration due to gravity changed to 19.6 m/s 2 ? Explain your reasoning. The period and frequency of a mass attached to a spring are independent of the acceleration due to gravity therefore the period and frequency remain the same

13. Question 6 B)What would happen to the frequency if the mass were quadrupled? Explain your reasoning.

14. Question 6 B)What would happen to the frequency if the mass were quadrupled? Explain your reasoning. T s = 2  m K

15. Question 6 b)The period would increase by the square root of 4 or double. The frequency would decrease by the square root of 4 or decrease by 1/2 T s = 2  m K

16. Question 7 What is the period of oscillation of a 1.0 kg mass attached to a 50.0 N / m spring that has been displaced 5.0 cm from its equilibrium position? T s = 2  m K

17. Question 7 What is the period of oscillation of a 1.0 kg mass attached to a 50.0 N / m spring that has been displaced 5.0 cm from its equilibrium position?.89 s = T s = 2  1.0 kg 50 N / m

18. Question 8 a) A guitar string has linear density of.020 kg / m. The tension on the guitar string is 20.0 N what is the velocity of a wave on the string? v = F T = 20 N ( m / L) (.020 kg / m) v = 32 m/s

19. Question 8 b) If the string with the linear density of.020 kg / m under 20.0 N is.75 meters long what is the lowest frequency that this string can play? v= f Fundemental L = ½ v = f 2 L

20. Question 8 b) If the string with the linear density of.020 kg / m under 20.0 N is.75 meters long what is the lowest frequency that this string can play? v = f = 32 m/s = 21 Hz 2L 2 (.75m)

21. Question 9 f = 256 Hz  = 2 L = 1.60m V = 409.6 m/s f = 512 Hz =2x256 Hz  = L =.80 m V = 409.6 m/s f = 768 Hz = 3x256 Hz  = 2 / 3 L =.53 m V = 409.6 m/s f = 1024 Hz=4X256 = 1/ 2 L =.40 m V = 409.6 m/s L =.80 m

22. Question 10 a) Describe or diagram a transverse wave. Displacement is perpendicular to wave propagation b) Describe or diagram a longitudinal wave. Displacement is parallel to wave propagation

23. Question 10 c) What type of wave is a water wave? transverse d) What type of wave is a sound wave? longitudinal e) What type of wave is a wave on a string? transverse

24. Question 10 e) v = 331 m/s +.6 T v= 331 m/s +.6 (20 o C) = 343 m/s Speed of sound is temperature dependent

25. Question 11a 11. The following open end tube has a length of.50m. a) What is the wavelength of this waveform? Since the tube is.50 m and the wave form represents a whole wave then the wavelength is.50 m ½ wave

26. Question 11 first b) What is the frequency of sound produced by this tube if the speed of sound is 345 m/s? V = f  v = f = 345 m/s = 690 Hz .50 m C) Which harmonic does this waveform represent? It represents the 2 nd harmonic of a open end tube

27. Question 11 first D)The following represents the 4 th harmonic of a open end tube ½ wave ½ wave ½ wave ½ wave

28. Question 11 #2 11. The following closed end tube has a length of.333 m. a) What is the wavelength of this wave form? This wave represents ¾ of a wave. Therefore L = ¾ or 4/3 L =  m ) =.444 m

29. Question 11 #2 b) What is the frequency of sound produced by this tube if the speed of sound is 345 m/s? v = f v = f = 345 m/s = 777 Hz.444 m

30. Question 11 #2 Which harmonic does this waveform represent? This represents the 3 rd harmonic of a closed end tube d) Draw the 1st harmonic or fundamental displacement wave form in the tube below ¼ wave ¼ wave ¼ wave

31. Question 11 #3a The relative velocity of the wave front determines the perceived frequency Stationary – source frequency = perceived frequency Source moving towards stationary observer- source frequency > perceived frequency Observer moving towards stationary source- source frequency > perceived frequency Observer and source moving towards each other source frequency > perceived frequency

32. Question 11 #3a The relative velocity of the wave front determines the perceived frequency Stationary – source frequency = perceived frequency Source moving away from stationary observer- source frequency < perceived frequency Observer moving away from stationary source- source frequency < perceived frequency Observer and away from each other source frequency < perceived frequency

33. Question 11 #3b Assume velocity of sound is 343m/s at 20 0 C f = 500 Hz ( 343m/s + 20 m/s )= 580 Hz ( 343 m/s – 30 m/s )

34. Tacoma Narrows Bridge Breaking Glass Comparison The Tacoma narrows bridge disaster was caused by standing waves produced in its structure by the wind which caused the bridge to vibrate at its natural frequency. Since the wind was applying a force at the natural frequency of the bridge, constructive interference occurred until the elastic limit of the materials the bridge was composed of was reached.

35. 13. Tacoma Narrows Bridge Breaking Glass Comparison Glass can be made to break if the glass is exposed to its natural frequency from a speaker or person at a very high amplitude. The sound from the person or speaker will cause the beaker to undergo constructive interference ( the glass will resonate ) until the elastic limit of the glass is met and it breaks.