Question

A certain organ pipe, open at both ends, produces a fundamental frequency of 285 Hz in air. Part A If the pipe is filled with helium at the same temperature, what fundamental frequency fHe will it produce? Take the molar mass of air to be 28.8 g/mol and the molar mass of helium to be 4.00 g/mol . Express your answer in hertz. View Available Hint(s)

Answer 1

Answer:

Explanation:

In an open organ pipe in fundamental mode of vibration

wave length of wave λ = 2L

where L  is length of  the pipe

frequency   = velocity of sound / λ

285 = 340 / 2 L

L = 59.65 CM

If the pipe is filled with helium at the same temperature, there will be change in the velocity of sound .So frequency of note produced will be changed .

velocity of sound is inversely proportional to  square root of molar mass of gas

velocity of sound in air / velocity of sound in helium = √4 / 28.8

340 / v =  0. 3725

v   =  912.75  m /s  

frequency

  = velocity of sound / λ

= 912.75 / (2 x . 5965)

= 765 Hz

Answer 2

The fundamental frequency of the sound in helium is 766.13 Hz.

What is the frequency?

The frequency is defined as the number of waves that pass a fixed place in a given time interval.

Given that a certain organ pipe, open at both ends, produces a fundamental frequency of 285 Hz in air.

Let's consider that the length of the pipe is L, then its wavelength is given as below.

$\lambda = 2L$

The frequency of the sound is given below.

$f = \dfrac {v}{\lambda}$

Where v is the velocity and f is the frequency of the sound in air.

$285 = \dfrac {331.29 }{2 L}$

$L = 0.5812 \;\rm m$

If the pipe is filled with helium at the same temperature, there will be a change in the velocity of sound. The velocity of sound is inversely proportional to the square root of the molar mass of gas.

If v' is the velocity of sound in helium, then

$\dfrac {v}{v'} = \sqrt{\dfrac {m'}{m}}$

Where m is the molar mass of air to be 28.8 g/mol and m' is the molar mass of helium to be 4.00 g/mol.

$\dfrac {331.29}{v'} = \sqrt{\dfrac {4}{28.8}}$

$v' = \dfrac {331.29}{0.372}$

$v' =890.56 \;\rm m/s$

Now the fundamental frequency of sound in helium is,

$f' = \dfrac {v'}{\lambda}$

$f' = \dfrac {890.56}{2\times 0.5812}$

$f' = 766.13 \;\rm Hz$

Hence we can conclude that the fundamental frequency of the sound in helium is 766.13 Hz.

To know more about the frequency, follow the link given below.

https://brainly.com/question/4393505.