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Math Help - difference between two frequencies of cellphone.

  1. #1
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    difference between two frequencies of cellphone.

    Two radio waves are used in the operation of a cellular telephone. To receive a call, the phone detects the wave emitted at one frequency by the transmitting station or base unit. To send your message to the base unit, your phone emits its own wave at a different frequency. The difference between these two frequencies is fixed for all channels of cell phone operation.

    Suppose the wavelength of the wave emitted by the base unit is 0.34341 m and the wavelength of the wave emitted by the phone is 0.36125 m. Using a value of 2.9979 x 10^8 m/s for the speed of light,

    determine the difference between the two frequencies used in the operation of a cell phone.


    I know that c or 2.9979 x 10^8 = frequency * wavelength

    however, i tried taking the difference of .34341 m and .36125m and then dividing c to retrieve an answer and that was incorrect. I know that is not the way this is calculated.

    Please show the conclusion please.
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  2. #2
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    Quote Originally Posted by rcmango View Post
    Two radio waves are used in the operation of a cellular telephone. To receive a call, the phone detects the wave emitted at one frequency by the transmitting station or base unit. To send your message to the base unit, your phone emits its own wave at a different frequency. The difference between these two frequencies is fixed for all channels of cell phone operation.

    Suppose the wavelength of the wave emitted by the base unit is 0.34341 m and the wavelength of the wave emitted by the phone is 0.36125 m. Using a value of 2.9979 x 10^8 m/s for the speed of light,

    determine the difference between the two frequencies used in the operation of a cell phone.


    I know that c or 2.9979 x 10^8 = frequency * wavelength

    however, i tried taking the difference of .34341 m and .36125m and then dividing c to retrieve an answer and that was incorrect. I know that is not the way this is calculated.

    Please show the conclusion please.
    Let l denote to the wavelength and f to the frequency.

    Since l_1<l_2~\implies~f_1>f_2

    Using your formula you get:

    f_1 = \frac c{l_1} .......... and

    f_2 = \frac c{l_2}

    And therefore:

    f_1 - f_2 = \frac c{l_1} - \frac c{l_2} = c \cdot \left(\frac1{l_1} - \frac1{l_2} \right) = c \cdot \left(\frac{l_2 - l_1}{l_1 \cdot l_2}\right)
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  3. #3
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    Hey thanks, I couldn't get the equation correct for this one, i knew I was to take the difference, then I wasn't sure if it was going to be negative if I had the numbers subtracted wrong.
    Awesome explanation.
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