110309

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course Phy 242

4/30 8

110309 Physics II`q001. The aluminum bar was 5 feet long and, with a node at the middle and antinodes at the ends, was observed to produce sound with 10 peaks in 6.1 milliseconds +- .1 millisecond.

What is the frequency of that sound?

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10peaks = 10cycles, which happens in approx. 6.1millisec = .0061sec

10cycles/.0061sec = approx. 1639CL/s

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What is its wavelength in the bar?

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5ft = (1/2)`lambda, `lambda = 10ft but only the 5ft is contained in the bar.

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What is the speed of propagation of sound in aluminum?

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c = freq*`lambda = 1639CL/s*10ft/CL = approx. 5000m/s or approx.

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What would be the wavelength of a sound of that frequency in air?

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speed of propagation of sound in air is approx. 340m/s, for freq = 1639CL/s

`lambda = 340m/s/1639CL/s = approx. .207m/CL

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`q002. For the wavelength you found in the preceding, for the sound as it propagates through air, what lengths of an open tube would you expect to produce resonance?

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.3105m, .5175m, .7245m, (n+.5)*.207m where n is +- integer.

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An open tube resonates in node-antinode configurations A N A, A N A N A, A N A N A N A, etc., corresponding to 1/2, 1, 3/2, ... wavelengths. Assuming wavelength .22 meters, to the length of the tube could therefore be

1/2 wavelength = 1/2 * .22 m = .11 m

1 wavelength = .22 m

3/2 wavelength = 3/2 * .22 m = .33 m

etc..

The effective length of the resonating air solum is in fact a bit greater than the length of the tube, but resonance should occur with tube lengths close to the calculated lengths.

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`q003. Where would you position supports in order to produce the second harmonic in the aluminum bar, and what would be the frequency of that harmonic?

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Supports would be at positions at (1/4) and (3/4)`lambda = 5/4m and 15/4m(because we orig had (1/2) a wavelength in rod. For antinodes at (1/4) and (3/4) we form a entire wave)

It would then cut our freq in half right, because we would be completing twice the cycles as before.????

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Where would you position supports in order to produce the third harmonic in the aluminum bar, and what would be the frequency of that harmonic?

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(1/6), (3/6) and (5/6) `lambda, for same reasons. Freq would be 3 times the original freq.

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What is the highest harmonic that would produce a sound audible to a healthy adult ear which is capable of hearing frequencies up to about 12 000 Hz? Where would you position supports in order to produce this harmonic?

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We would solve for

n*(1639Hz) = 12000Hz

n = approx. 7.3215, so 7 would be the highest harmonic.

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The first few harmonics correspond to node-antinode configurations A N A, A N A N A, A N A N A N A.

The supports need to be at nodes.

The node-antinode configuration of the second harmonic divides the bar into four equal sections. The nodes occur at 1/4 of the bar length from each end. For a 5-foot bar, the supports would therefore be 5/4 ft from each end (1 ft and 3 inches, or 15 inches, from each end).

The third harmonic divides the bar into 6 equal node-antinode distances, so supporta 5/6 ft from each end (10 inches) would be expected to produce the third harmonic.

With a propagation speed of about 5 000 m/s, a wave 5/12 meter long would be expected to produce a 12 000 Hz sound. Thus the 1/4 of a cycle between an antinode at the end and the adjacent node would correspond to 1/4 * 5/12 meter = 5/48 meter, about 10.2 cm.

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`q004. The wave below represents a sound wave, with the vertical axis representing pressure relative to atmospheric pressure in units of Pascals, and the horizontal axis representing clock time in milliseconds.

What is the frequency of this wave?

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2sec for one cycle to complete, cycle consists of 2`pi radians so

freq. = 2`pi rad/2millisec = `pi rad/millisecond, which is (1/2) cycle/millisecond

I’m pretty sure I got the logic (should by now, right) but when we use the formula are we supposed to know that 2`pi = 1 cycle and `pi = (1/2) cycle, etc.. or am I forgetting something in my formula that cancels the radian units out???

Found my errors

freq. = 2`pi rad/2millisec = `pi rad/millisecond = 500cycle/sec = 500Hz

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If sound travels at 340 m/s, what would be its wavelength in air?

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`lambda = c/freq. =(340 m/s)/ (1/2)cycle/millisec) = (340000m/millisec)/((1/2)cycle/millisec)

= 680000m/cycle

?????I think I’m right but seem high, let me know if I missed something????????

I left my previous answer because I didn’t see my error till the last question, but I understand my freq. should be in cycles per sec not cycles per millisec

`lambda = c/freq. =(340 m/s)/ (500Hz) = .68m/cycle

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If this sound is produced as the fundamental harmonic in an aluminum bar whose ends are free, what is the length of the bar?

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L = `lambda/2, for fundamental freq.

L = (.68m/cycle)/2 = .34m or 340cm

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From peak to peak we have a change of 2 in the horizontal coordinate, corresponding to 2 milliseconds. The wave therefore has period 2 milliseconds = .002 seconds, and frequency 1 / (.002 sec) = 500 Hz.

At 340 m/s, the wavlength of a 500 Hz wave would be (340 m/s) / (500 cycles / sec) = .68 meters.

In aluminium, where propagation speed is around 5 000 m/s, the wavelength would be 10 meters. The fundamental harmonic has configuration A N A, corresponding to half a wavelength, so the wavelength is double that of the bar. The bar would therefore be about 5 meters long.

The higher-pitched wave in the second figure has period .5 millisecond and hence frequency 2000 Hz. In air this wave would have wavelength 340 m/s / (2000 cycles / sec) = .17 meter, or 17 cm.

As the waves are depicted, the 'peaks' of the first wave all align with 'valleys' of the second, which results in the indentation at the 'top' of the combined wave; note the symmetry of that indentation. The 'valleys' of the first wave also align with 'valleys' of the second, and the two 'valleys' reinforce, causing a deeper 'valley' for the combined wave.

If the second wave is shifted slightly to the right, then the indentation at the top of the combined wave would be shifted slightly to the right. The symmetry of that indentation would be lost, with the peak just to the left of the indentatation becoming a little higher and the peak to the right becoming a little lower. The 'valleys' of the two waves would no longer coincide so the 'valley' of the combined wave would not be quite as deep, and the shape of the wave to the left would no longer be a reflection of the shape to the right.

If the second wave is shifted a little further to the right the peaks of the first could align with the peaks of the second, causing a higher peak in the combined wave. In this case the valleys of the first would align with the peaks of the second, and the valleys of the combined wave would have upward 'humps'. The combined wave would be the same as the one depicted above, but flipped upside down.

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&#Your work looks good. See my notes. Let me know if you have any questions. &#