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Question

In Fig. S is a small loudspeaker driven by an audio oscillator with a frequency that is varied from 1000 Hz to 2000 Hz, and D is a cylindrical pipe with two open ends and a length of 45.7 cm. The speed of sound in the air-filled pipe is 344 m/s. (a) At how many frequencies does the sound from the loudspeaker set up resonance in the pipe? What are the (b) lowest and (c) second lowest frequencies at which resonance occurs?
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Solution

(a) Each block is assumed to have uniform density, so that the center of mass of each block is at its geometric center (the positions of which are given in the table [see problem statement] at t=0 ). Plugging these positions (and the block masses) into readily gives xcom=0.50m (at t=0 )
(b) Note that the left edge of block 2 (the middle of which is still at x=0) is at x=2.5cm , so that at the moment they touch the right edge of block 1 is at x=2.5cm and thus the middle of block 1 is at x=5.5cm. Putting these positions (for the middles) and the block masses into Eq. 929 leads to xcom=1.83cm or 0.018m (at t=(1.445m)/(0.75m/s)=1.93s) .
(c) We could figure where the blocks are at t=4.0s again, but it is easier (and provides more insight) to note that in the absence of external forces on the system the center of mass should move at constant velocity:

νcom=m1ν1+m2ν2m1+m2=0.25m/s^i
as can be easily verified by putting in the values at t=0 . Thus ,
xcom=xcom initial+νcomt=(0.50m)+(0.25m/s)(4.0s)=+0.50m.

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