If one holds an open-open tube with one end in water, one will obtain a variable

a) lambda = 4 L = 4 * 0.09 = 0.36 f = lambda f = 0.36 * 343 = 123.48 = 123 Hz b) L2 = 3 L = 3 * 0.09 = 0.27 m L3 = 5 L = 5 * 0.09 = 0.45 m

ID: 2276792 • Letter: I

Question

If one holds an open-open tube with one end in water, one will obtain a variable length open-closed

tube, the length depending on how much of the tube is immersed. Then if one holds a tuning fork over the open end

of the tube, one will obtain resonances when the length, L, of the tube is such that the sound of the tuning fork will

be reinforced. The smallest value of L for which a resonance occurs is 9.0 cm. What is the frequency of the tuning

fork, and what are the next two values of L for which resonance will occur? Use vs = 343 m/s. (953 Hz, 0.27 m,

0.45 m)

lambda = 4 L = 4 * 0.09 = 0.36

f = lambda f = 0.36 * 343 = 123.48 = 123 Hz

L2 = 3 L = 3 * 0.09 = 0.27 m

L3 = 5 L = 5 * 0.09 = 0.45 m

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