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Svet_ta [14]
3 years ago
9

A cylindrical tube sustains standing waves at the following frequencies: 600 Hz, 800 Hz, and 1000 Hz. The tube does not sustain

standing waves at 500 Hz, at 900 Hz, at any frequencies between 600 and 800 Hz, or at any frequencies between 800 and 1000 Hz. Determine the fundamental frequency of the tube and whether the tube is open at both ends or has only one end open
Physics
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

Explanation:

In  closed organ pipe notes with odd harmonics are produced and in open organ pipe notes with all odd and even harmonics are produced. notes with frequencies 600, 800 and 1000 Hz are produced. These are 3 , 4 and 5 times 200 Hz. ie both odd and even times of 200 . So fundamental frequency appears to be 200 Hz. There is no note available between 800 and 1000. It also indicates that 200 Hz is the fundamental frequency and the pipe is open at both ends.

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<em>We have a formula for such condition,</em>

Q=m.c.\Delta T.....................................(1)

where:

  • \Delta T= temperature difference
  • Q= heat energy
  • m= mass of the body
  • c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

  • we have equal masses of two solutions A & B, i.e. m_A=m_B
  • equal heat is supplied to both the solutions, i.e. Q_A=Q_B
  • specific heat of solution A, c_{A}=2.0 J.g^{-1} .\degree C^{-1}
  • specific heat of solution B, c_{B}=3.8 J.g^{-1} .\degree C^{-1}
  • \Delta T_A & \Delta T_B are the change in temperatures of the respective solutions.

Now, putting the above values

Q_A=Q_B

m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1

Which proves that solution A attains a higher temperature than solution B.

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