Physics LO 101 Beats

Q. Ed, a friend of yours, has a tuning company and is

receiving numerous complaints on a new employee. He asks you to tag along during the next tune up to find out what the problem is. During the tune up the employee is unable to determine the frequency of a tuning fork. He asks for your help and tells you the following things:

The beat frequency between a 702 Hz

tuning fork and an unknown tuning fork is 20 Hz

With an 708 Hz tuning fork the beat frequency is 26 Hz

Find the frequency of the unknown tuning fork.

Solution After applying the principle of superposition and adding two waves the expression given is:

stotal(0,t) = 2sm cos(-ϖt)cos(Δωt)

ϖ = (ω1 + ω2)/2 Mean angular frequency

Δω = (ω1 - ω2)/2 Angular frequency difference Although the difference of frequency is defined as Δω = (ω1 - ω2)/2 for the wave. What is heard is by our ears are the two individual frequencies that make up the wave. However, since the wave is getting louder and softer at the frequency of (ω1 -

ω2)/2 it results in us hearing “beats”. The beat frequency is given by the absolute value of the two slightly different frequencies. Therefore, If x = the unknown frequency, then

20 Hz = | x-702| Hz (1) 26 Hz = |x-708| Hz (2)

Both of these have two solutions, In (1) the unknown frequency could be 722 Hz or 682 Hz In (2) the unknown frequency could be 734 Hz or 682 Hz We know that the beat frequency increased from 20 to 26 Hz as the known frequency increased from 702 to 708 Hz. So, in order to distinguish between these answers, the unknown frequency must be lower than 702 Hz and satisfy both equations. The frequency of the unknown tuning fork = 683 Hz