Sample problem: A piano string is 1.1 m long and has a mass of 9 g. How much tension must the string be under if it is to vibrate at a fundamental frequency of 131 Hz? What are the frequencies of the first few harmonics?
Solution:


 

Sample problem: In an experiment on standing waves, a string 90 cm long is attached to the prong of an electrically driven tuning fork that oscillates perpendicular to the length of the string at a frequency of 65 Hz. The mass of the string is 0.055 kg. What is the mass of the suspended weight if the string vibrates in four loops?
Solution: The key here is that the weight of the suspended mass equals the tension, and it is the tension which controls the velocity of the wave on the string.




Sample problem: One of the harmonics on a 1.3-m-long string has a frequency of 15.6 Hz. The next higher harmonic has a frequency of 23.4 Hz. What is the fundamental frequency? What is the speed of the wave?
Solution:




Sample problem: A hollow tube is partly submerged in water, as shown. If the smallest value of L for which a peak occurs is 9 cm, what is the frequency of the tuning fork and the value of L for next two resonant modes?
Solution:

f1 = v/4L = 953 Hz
L2 = 3l/4 = 0.27 m
L3 = 5l/4 = 0.45 m


 

Sample problem: A tuning fork is set into vibration above a vertical open tube filled with water. The water level is allowed to drop slowly. As it does so, the air in the tube above the water level is heard to resonate with the tuning fork when the distance from the tube opening to the water level is 17.5 cm, 53.5 cm, 89.5 cm, and 125.5 cm. If the speed of sound is 343 m/s, what is the frequency of the tuning fork? What is the end correction (i.e., the distance from the end of the tube to the node or antinode just outside the tube)?
Solution
: At resonance, the displacement antinode lies a little beyond the open end of the tube. This so-called end correction is typically around .6*R (R=radius) for a tube with a circular cross-section.


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