Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air to 8 m underwater. How has the change in media affected the frequency of the wave?

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skyluke89 skyluke89 · Answer 1 · 2018-10-10T11:17:16+02:00

The frequency of the wave has not changed.

In fact, the frequency of a wave is given by:

[tex]f=\frac{v}{\lambda}[/tex]

where v is the wave's speed and [tex]\lambda[/tex] is the wavelength.

Applying the formula:

- In air, the frequency of the wave is:

[tex]f=\frac{400 m/s}{2 m}=200 Hz[/tex]

- underwater, the frequency of the wave is:

[tex]f=\frac{1600 m/s}{8 m}=200 Hz[/tex]

So, the frequency has not changed.

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xero099 xero099 · Answer 2 · 2021-11-29T13:58:57+01:00

The change in media affected the frequency by a quarter.

The sound velocity ([tex]v[/tex]), in meters per second, is described by the following expression:

[tex]v = \lambda\cdot f[/tex] (1)

Where:

By (1), we get expression for the frequency:

[tex]f = \frac{v}{\lambda}[/tex]

Now we proceed to determine the frequency associated to each medium:

Water ([tex]v = 400\,\frac{m}{s}[/tex], [tex]\lambda = 8\,m[/tex])

[tex]f = \frac{400\,\frac{m}{s} }{8\,m}[/tex]

[tex]f = 50\,hz[/tex]

Air ([tex]v = 400\,\frac{m}{s}[/tex], [tex]\lambda = 2\,m[/tex])

[tex]f = \frac{400\,\frac{m}{s} }{2\,m}[/tex]

[tex]f = 200\,hz[/tex]

The change in media affected the frequency by a quarter.

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