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2 years ago

An open-pipe resonator has a length of 2.39 m. Calculate the frequency of its third harmonic if the speed of sound is 343 m/s.

Physics

1 answer:

Verdich [7]2 years ago

8 0

Answer:

f = 215.27 Hz

Explanation:

We have,

An open-pipe resonator has a length of 2.39 m.

It is required to find the frequency of its third harmonic if the speed of sound is 343 m/s.

For third harmonic, its length is equal to, $l=\dfrac{3\lambda}{2}$

So,

$\dfrac{3v}{2f}=2.39$

f is frequency in its third harmonics

$f=\dfrac{3v}{2\times 2.39}\\\\f=\dfrac{3\times 343}{2\times 2.39}\\\\f=215.27\ Hz$

So, the frequency of its third harmonic is 215.27 Hz.

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Answer:

We can retain the original diffraction pattern if we change the slit width to d) 2d.

Explanation:

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Using the following parameters: y for the distance from the center of the bright maximum to a place of minimum intensity, m for the order of the minimum, λ for the wavelength, D for the distance from the slit to the screen where we see the pattern and d for the slit width. The distance from the center to a minimum of intensity can be calculated with:

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3 years ago

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Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
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