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4[25] Your lab TA gives you a thin, uniform, air-filled tube with a speaker at one end. A signal generator that can be tuned to emit any frequency between 150 Hz and 750 Hz by turning a dial drives the speaker.

v_air = 340 m/s

a[20] With the far end (away from the speaker) closed, you observe by reading the dial that the following frequencies are amplified: 296 Hz, 507 Hz and 705 Hz. (SHOW ALL WORK, an answer with no work will not be accepted)

i[8] What is the fundamental frequency of the closed tube?

A closed tube resonates for:   _{n = (nvair)/(4L), where n=1,3,5,…. Observed frequencies are in ratio 3:5:7. We therefore are missing the fundamental frequency that we infer to be 100 Hz.}

n₁ = 100 Hz

ii[5] What is the length of the tube? L = v_air/(4 n₁)

L = (340m/s)/4(100Hz) = 0.85 m

iii[7] Draw in the tubes below the standing displacement wave for the two lowest resonant frequencies you observe and indicate clearly the position of the displacement nodes and antinodes in meters.

Displacement nodes/antinodes coincide with pressure antinodes/nodes respectively.

b[5] What are the three lowest resonant frequencies of the same tube with the far end open?

An open tube’s resonant frequencies are given by:   _{n = (nvair)/(2L), where n + 1,2,3,…}

_{Consequently,}   n_{1 =} (340m/s)/2(0.85m) = 200 Hz, n_{2 = 2}n_{1   = 400Hz,} n_{3 = 3}  n_{1 = 600 Hz}