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svet-max
9 days ago
13

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? It increased. It decreased. It stayed the same. It cannot be calculated.
Physics
2 answers:
Sav [2.8K]8 days ago
8 0

Response: Frequency remains unchanged.

Clarification:

Voices conveyed through air travel at 400 m/s, while underwater they reach speeds of 1600 m/s.

The wavelength shifts from 2 m in air to 8 m in water.

The relationship between speed and wavelength reveals that

v=\nu\times \lambda

defines frequency.

\nu=\dfrac{v}{\lambda}

\nuis the wavelength.

\lambdaFrequency through air,

Frequency in water, \nu_a=\dfrac{400\ m/s}{2\ m}=200\ Hz

\nu_w=\dfrac{1600\ m/s}{8\ m}=200\ HzConsequently, the frequency of the swimmers' voices does not change throughout the process.

Thus, the correct choice is (C) "It stayed the same".

Keith_Richards [2.9K]8 days ago
5 0

The frequency of a sound remains constant as it departs from the source. It does not alter.

The voices of swimmers do not modify in frequency when transitioning to or from the water. Only their speed and wavelength vary.

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