# Math Help - difference between two frequencies of cellphone.

1. ## 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.

2. Originally Posted by rcmango
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.

Let l denote to the wavelength and f to the frequency.

Since $l_1f_2$

$f_1 = \frac c{l_1}$ .......... and
$f_2 = \frac c{l_2}$
$f_1 - f_2 = \frac c{l_1} - \frac c{l_2} = c \cdot \left(\frac1{l_1} - \frac1{l_2} \right) = c \cdot \left(\frac{l_2 - l_1}{l_1 \cdot l_2}\right)$