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Question: (II) (a) Determine the length of an open organ pipe that emits middle C (262 Hz) when the temperature is 18°C. (b) What are the wavelength and frequency of the fundamental standing wave in the tube? (c) What are \(\lambda \)and f in the traveling sound wave produced in the outside air?

Short Answer

Expert verified
  1. The length of the pipe is \(0.652{\rm{ m}}\).
  2. The frequency and the wavelength of the standing wave are \(262{\rm{ Hz}}\) and \(1.30{\rm{ m}}\).
  3. The frequency and the wavelength in the outside air is also \(262{\rm{ Hz}}\) and \(1.30{\rm{ m}}\).

Step by step solution

01

Step-1:-Relation of fundamental frequency

In this problem, for evaluating the length of an open organ pipe the relation of fundamental frequency of an open pipe will be used.

02

Step-2:-Given data

The frequency of the pipe is \({f_1} = 262{\rm{ Hz}}\).

The given temperature is \(T = 18{\rm{ }}^\circ {\rm{C}}\).

03

Step-3:-Calculation of the length of the pipe

The speed of sound at the temperature \(18^\circ {\rm{C}}\) is \(\nu = 341.8{\rm{ m/s}}\).

The length of the open pipe is calculated as,

\(l = \frac{\nu }{{2{f_1}}}\)

Substitute the values in the above relation.

\(\begin{array}{c}l = \frac{{341.8{\rm{ m/s}}}}{{2 \times \left( {262{\rm{ Hz}}} \right)}}\\l = 0.652{\rm{ m}}\end{array}\)

Thus, the length of the pipe is \(0.652{\rm{ m}}\).

04

Step-4:-Calculation of the wavelength of the standing wave

The frequency of the standing wave in the pipe is \(262{\rm{ Hz}}\).

The wavelength of the standing wave is calculated as,

\(\lambda = 2l\)

Substitute the values in the above relation.

\(\begin{array}{c}\lambda = 2 \times 0.652{\rm{ m}}\\\lambda = {\rm{1}}{\rm{.30 m}}\end{array}\)

Thus, the wavelength and frequency is \(1.30{\rm{ m}}\) and \(262{\rm{ Hz}}\).

05

Step-5:-Calculation of the frequency and wavelength in the outside air

The frequency and wavelength of the traveling sound wave produced in the outside air will be the same as the value of frequency and wavelength in the organ pipe because the same air is in present both conditions.

Thus, the frequency is \(262{\rm{ Hz}}\) and the wavelength is \(1.30{\rm{ m}}\).

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