TBR Physics 6-26

[Reine]

Hello,

I've just gone over some of these TBR physics questions, but this one has me stuck. The question is:

"In Trial 5, the height of the water was 75.0 cm. The tuning fork from which other trial could be used to produce a resonance for a water height of 75.0 cm? (the total tube length was 1m)"

a. Trial 1 - Frequency = 1020 Hz, water height = 91.7 cm
b. Trial 2 - Frequency = 850 Hz, water height = 90.0 cm
c. Trial 3 - Frequency = 680 Hz, water height = 87.5 cm
d. Trial 4 - Frequency = 510 Hz, water height = 83.3 cm

The answers says it's trial 1, which has 3x the frequency given the same amount of air in the tube... However, I don't see how it's a better answer than c, which would be 2x the frequency.

I'd be grateful if someone could explain this.
(And also, if Berk ever sees this - I think your last paragraph on p 13 has contradictory info to the chart on the next page with regards to the vs signs... unless I'm getting this all wrong :s)

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[Majik]

First, you have to find out how much free space is in the tube:

The length of the pipe = 1m (100cm)
The height of the water in the pipe = 75cm
Therefore the free space = 25cm (100cm-75cm)

The "Length" for the standing wave is 25 cm (or 0.25m).

Because it's closed at one end, the equation we use is: frequency: nv / 4L
"v" equals the speed of the wave in air: 340m/s.

We can simplify this down to:
frequency = n(340m/s)/4(1/4m)
frequency = (340Hz)n

Since the pipe is open at one end, only odd harmonics can be used.

Therefore, we need a number that is some odd variable of 340 Hz.

340 x 1 = 340 Hz <--- Trial 5
340 x 3 = 1020 Hz <--- Trial A!!
340 x 5 = 1360 Hz

Trial A also has a harmonic frequency under those conditions, which is why it's the answer.

 

[Reine]

Oooh, I see! I forgot that only odd harmonics worked in this case.

Thank you very much Majik!

 

[geeyouknit]

Apologies for reviving this thread, but can someone explain why we only use the air / free space (25 cm) and not the water (75 cm)?

Thanks.

 

[circulus vitios]

Apologies for reviving this thread, but can someone explain why we only use the air / free space (25 cm) and not the water (75 cm)?

Thanks.

In the problem it says L is the length of the air column. Beyond that I have no idea.

 

[Majik]

Apologies for reviving this thread, but can someone explain why we only use the air / free space (25 cm) and not the water (75 cm)?

Thanks.

Here's something I encountered in a TBR passage: "Another law of wave motion deals with how sound waves reflect when they hit an interface between two different media. The ratio of the reflected wave's intensity to the incident wave's intensity is given by: [insert: (some equation you don't need to know)]." Basically, the more different the densities between two media, the more waves will reflect. Air and Water have two different densities.

 

[RiPSTONE]

Thanks @Majik for the explanation!