TPRH Workbook #346 Waves

[BeatMCAT]

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So, the question is a 1.5 m long organ pipe, closed at one end, is resonating in third harmonic, with wavelength lamda 3. How does this resonant wavelength compare to the fundamntal resonant wavelength lamda 1?

a) Lambda 3= 1/3 lambda1 (ANSWER apparently)
b) lambda 3 = 2/3 lambda 1
c) lambda 3= 3/4 lambda 1
d) lambda 3= 3 lambda 1.

So, the fundamental wavelength has to be 1.5 * 4 = 6m. so I thought the third harmonic wavelength would be (4/5)L, which adds up to 1.2. Now how in the hell is wavelngth 1.2 = 6/3. Am i wrong or TPRH made a mistake. Thanks!!!

 

[Majik]

I had this confusion too and I know exactly what you're line of thinking was.

You're right that the harmonic series for an open pipe allows only odd numbers only (1,3,5,7,etc.). However, the third harmonic is 3 just as the fifth harmonic is 5. I think your confusion was mistaking the third harmonic to be 5 which is NOT true.

Secondly, since the third harmonic is 3 this means that a wave oscillating at its third harmonic will be three times smaller than the fundamental wavelength:

4L/3 : 4L/1 ---> 1:3 or Wavelength (3rd harmonic) : 3 x Wavelength (1st Harmonic)

This is why the answer is A.

 

[BeatMCAT]

I had this confusion too and I know exactly what you're line of thinking was.

You're right that the harmonic series for an open pipe allows only odd numbers only (1,3,5,7,etc.). However, the third harmonic is 3 just as the fifth harmonic is 5. I think your confusion was mistaking the third harmonic to be 5 which is NOT true.

Secondly, since the third harmonic is 3 this means that a wave oscillating at its third harmonic will be three times smaller than the fundamental wavelength:

4L/3 : 4L/1 ---> 1:3 or Wavelength (3rd harmonic) : 3 x Wavelength (1st Harmonic)

This is why the answer is A.

Ohhh I see. Thanks a lot! Appreciate it.