A learning object that explains the concept of how constructive and destructive interference between two sound waves with varying frequncies produces beats. Practice problems with answers are included to improve understanding.

1. SOUND
INTERFERENCE:
BEATS
Leslie Liu
Learning Object

2. Interference
 Waves undergo constructive and destructive
interference
 Constructive Interference: when two waves are
in phase at a certain time
 Eg. Two waves are at their maximum positive
amplitude
 Destructive Interference: when two waves are
completely out of phase at a certain time
 Eg. One wave is at maximum positive amplitude
and the other is at maximum negative amplitude

3. Interference

4. Interference
 Two waves with slightly different frequencies
will periodically be in phase then out of phase
 A maximum amplitude results when the
crest/trough of the waves coincide
 Maximum sound level is heard (max intensity)
 Sound level decreases as the waves move out
of phase
 A minimum amplitude results when the crest of
one and the trough of the other wave coincides
 Minimum sound level is heard (min intensity)

5. Interference
The two waves move periodically in and out of
phase

6. Interference
 The amplitude of the resultant wave increase
and decrease as they undergo constructive
and destructive interference
 The pattern of constructive and destructive
interference between two sound waves
produces beats

7. Interference
The pattern of interference produces beats

8. Beats
 Beats: the variation in amplitude that occurs
between two waves with different frequencies
 Beat Frequency: the number of times the
amplitude (sound level) reaches a maximum in
a second
 the difference of the two frequencies

9. Beats
 The sound we hear is the average of the two
frequencies
 When the difference in frequency is large, we
hear two distinct tones
 When the difference in frequency is small, we
hear one tone that beats (varies in intensity)

10. Question 1:
1) You tune your guitar with an electric tuner
that emits the note “A” at a frequency of 440
Hz. Your A string is out of tune and has a
frequency of 444 Hz. What do you hear when
you pluck the A string to tune your guitar?
A) A tone at 442 Hz and 2 beats per second
B) A tone at 884 Hz and no beats per second
C) A tone at 442 Hz and 4 beats per second
D) A tone at 884 Hz and 4 beats per second

11. Question 1: Answer
 C) A tone at 442 Hz and 4 beats per second
 The resultant tone we hear is the average of
the two frequencies
 (440 Hz+444 Hz)/2 = 442 Hz
 The number of beats per second is determined
by finding the difference in frequency between
the two sounds
 |440 Hz – 444 Hz|= 4 beats per second

12. Question 2:
2) Two trombone players are tuning before a
concert. Alvin plays a note with a frequency of
466 Hz. Bob plays the same note, but his note
is flat and beats occurring 5.0 times per
second can be heard. Hearing this, Bob
adjusts his tuning slide before playing again.
This time only 2.0 beats per second can be
heard.
a) What is the original frequency of Bob’s
note?
b) What is the new frequency of Bob’s note?
c) If you are listening to them play the same
note after tuning, what do you hear?

13. Question 2: Answer Part A
 The beat frequency is the difference of the two
frequencies
 The frequencies differ by 5.0 Hz because beats
occur 5.0 times/second
 When a note is flat, it has a frequency that is
slightly lower than the original note
 Using the beat frequency equation:
 466 Hz – x = 5.0 beats/second
 x = 461 Hz

14. Question 2: Answer Part B
 Bob adjusts his tuning slide so there are now
fewer beats that occur
 The difference in the two frequencies has
decreased
 In order for this to occur, Bob must increase
his original frequency to match Alvin’s
 Using the beat frequency equation:
 466 Hz – x = 2.0 Beats/second
 x = 464 Hz

15. Question 2: Answer Part C
 Bob’s new frequency = 464 Hz
 Alvin’s frequency remains constant at 466 Hz
 Using the average frequency formula:
 favg = (464 Hz + 466 Hz)/2
 favg = 465 Hz
 After the adjustment, 2 beats/second occur
 You hear a note with a frequency of 465 Hz
and 2 beats/second

16. Question 3:
3) A violinist is tuning her instrument with a
tuning fork that emits a frequency of 659 Hz.
She strikes the tuning fork and plays a note
at the same time. The frequencies are similar,
but she hears 3.0 beats/second. She loosens
her violin string and plays again. She now
hears 2.0 beats per second.
a) What was her original frequency?
b) By what percentage did the tension
decrease?

17. Question 3: Answer Part A
 The frequencies differ by 3.0 Hz because beats
occur 3.0 times/second
 The violinist loosens the string (decreases tension)
 Decreases the speed of waves on the string
which decreases frequency
 Fewer beats heard after the adjustment means
we are moving closer to the frequency of the
tuning fork
 The original frequency must be higher than 659 Hz
 Original frequency = 662 Hz

18. Question 3: Answer Part B
 Recall the equation for wave speed
 The linear mass density does not change
 Recall another equation for wave speed
 Wavelength (λ) does not change
 λ = wavelength of the transverse wave on the
string
 Since λ doesn’t change, v must be proportional
to f
 Therefore the frequency is proportional to the
square root of tension

19. Question 3: Answer Part B
 This relationship can be rewritten as the
tension is proportional to the frequency
squared
 Set up a ratio to compare tension

20. Question 3: Answer Part B
 There are now fewer beats that occur
 The difference in the two frequencies has
decreased
 The original frequency must decrease to
become closer to the frequency of the tuning
fork
 New frequency = 661 Hz
 Ratio of tension:
 (661 Hz)2/(662 Hz)2 = 0.997
 Subtract from one to find the decrease in
tension
 1 – 0.997 = 0.003 = 0.3%

21.  Physics for Scientists and Engineers – An
Interactive Approach
 Images:
 http://hyperphysics.phy-astr.gsu.edu/hbase/sound/interf.html
 http://www.a-levelphysicstutor.com/wav-beats.php
 http://faculty.wcas.northwestern.edu/~infocom/Ideas/lightsound.ht
ml
Works Cited