Summary of beats with clicker questions

1. E D I E Y E
BEATS

2. BEATS
• Beats are caused by the interference of waves that
have different frequencies
• To find the resulting wave of the beat, we use the
principle of superposition to add the two waves

3. PRINCIPLE OF SUPERPOSITION
• To add the waves we use the formula:
s(0, t) = 2 sm cos(-ϖt)cos(Δϖt), where ϖ is the mean
angular frequency and Δϖt is the angular
frequency difference

4. CONNECTION
• What is a purpose of beats?
• A purpose of beats is the formation of music using
waves that have different frequencies. Similar to
how increasing the music on your iPod increases the
amplitude of the wave, the formation of the distinct
genres of music such as EDM Ii.e. electronic dance
music) comes from the different frequencies of the
waves interfering.

5. BEAT FREQUENCY
• The beat frequency is the frequency difference
between interfering waves
• When this frequency is large, we hear two distinct
tones but when it is small, we only hear one
combined wave

6. CLICKER 1
• The period of wave one is 0.002 seconds and the
period of wave two is 0.002008 seconds. Which of
the frequencies are the highest and what beat
frequency would we expect to hear if the waves
were played simultaneously?

7. CLICKER 1
• a. wave one; 2
• b. wave one; -2
• c. wave two; 2
• d. wave two; -2
• e. can’t tell from information given

8. CLICKER 1 EXPLAINED
• Even without calculation we can tell that frequency
of wave one has a higher frequency as period and
frequency are inversely proportional and wave one
has a shorter period. From the options available, we
can see that the frequency is either positive two or
negative two, but because frequency can not be
negative as it is function of time we can conclude
that the answer is 2, thus the final answer is a.
• Calculation of wave one: 1/ 0.002 = 500 Hz
• Calculation of wave two: 1/0.002008 =498 Hz
• Calculation of beat frequency: 500 Hz – 498 Hz =
2Hz

9. CLICKER 2
• A 300 Hz tone and a 304 Hz tone are played
simultaneously over a period of five seconds. How
many beats can we expect to hear?

10. CLICKER 2
• a. 4
• b. 8
• c. 12
• d. 16
• e. 20

11. CLICKER 2
• We first calculate the beat frequency which is
304 Hz- 300Hz =4Hz and then multiply by the
duration (i.e. 5 seconds) to get 20 beats over the
time frame.

12. CLICKER 3
• The beat frequency of a source is 1 Hz. The tone of
one source is 100 Hz. What can we expect the
frequency of the other source to be?

13. CLICKER 3
• a. 99
• b. 100
• c. 101
• d. 99 Hz or 101 Hz
• e. 100 Hz or 101 Hz

14. CLICKER 3 EXPLAINED
• Given the beat frequency of one Hz and the
frequency of 100 for one tone we can see that
there are two possible answers for the second tone.
This is because the frequency is an absolute value
and |99-100|would give us one Hz and |101-100 |
would give us one Hz as well thus the answer is d.

15. THANK YOU FOR WATCHING

16. WORKS CITED
• Hawkes, Robert, Javed Iqbal, Firas Mansour,
Marina-Bolotin, and Peter Williams. Physics for
Scientists and Engineers. Vol. 1. : Nelson, n.d.
Print.