alexyssewell
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Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.

1. wavelength = frequency / speed
2. speed = frequency x wavelength
3. wavelength = speed / frequency
4. frequency = speed / wavelength
5. frequency = wavelength over speed
6. speed = wavelength x frequency

Respuesta :

AL2006
-- Equations  #2  and  #6  are both the same equation,
and are both correct.

-- If you divide each side by  'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by  'frequency', you get Equation #3,
which is also correct. 
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.

The correct answers to the question are 2,3,4 and 6 options respectively.

EXPLANATION:

Let us consider a wave which is moving with a speed v in a medium .

Let f and [tex]\lambda[/tex] is the frequency and wavelength of the wave.

The relation between frequency,wavelength and speed is given as -

                       speed = [tex]=\ frequency\times wavelength[/tex]

                                   = [tex]f\times \lambda[/tex]

The wavelength of the wave is calculated as -

                             [tex]wavelength=\ \frac{speed}{frequency}[/tex]

Similarly frequency of the wave can be calculated as-

                            [tex]frequency=\ \frac{speed}{wavelength}[/tex]

The option 1 is a wrong relation. So, we can not calculate the wavelength through this equation.

The option 2 is a right relation. By rearranging it, we can calculate the frequency of the wave.

The option 3 is a correct relation. So, we can calculate the wavelength through it.

The option 4 is also a correct relation. By rearranging it, we can calculate the wavelength.

The option 5 is a wrong relation.

The option 6 is also a right relation. By rearranging it, we can calculate the wavelength of the wave.

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