SOUND INTERFERENCE:

BEATSLeslie Liu

Learning Object

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

Interference

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)

Interference

The two waves move periodically in and out of phase

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

Interference

The pattern of interference produces beats

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

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)

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 secondC) A tone at 442 Hz and 4 beats per secondD) A tone at 884 Hz and 4 beats per second

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

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?

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

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

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

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?

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

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

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

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%

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/light

sound.html

Works Cited