Physics 101 Learning Objective 7

Beats & Intensity Cindy Shen 03/11/15

What exactly are “beats”?

Beats are the periodic variation of sound levels that occur when 2 waves of slightly different frequencies interfere.

http://astarmathsandphysics.com/ib-physics-notes/waves-and-oscillations/ib-physics-notes-beats.html

Beats applied to musical instruments…

The resulting sound of beats is a single frequency sound that periodically varies in amplitude or loudness.

Beats are sometimes used by musicians to tune instruments. Beat frequency typically decreases when instruments are tuned, and they eventually disappear.

Guitar tuner used by musicians. http://www.qwiktunetuners.com/qwiktuners.html

Beats happens when…

• 2 waves with EQUAL AMPLITUDE but SLIGHTLY DIFFERENT FREQUENCIES overlap

• Creates wave of frequency

fav = (f1 + f2)/2 … which amplitude wavers with beat

frequency

fbeat = l f2 – f1 l

Question 1: Lisa and Ryan are playing a song together. Lisa is playing the violin at 196Hz while Ryan is playing the cello at 300Hz. Assuming the phase constant is 0, what is the mean angular frequency?

1. Instead of looking for the average frequency, the question asks for the mean (average) angular frequency. In the past, we learned ω = 2pif.

ω(Lisa) = 2pif. = 2pi(196Hz) = 1231.50 rad/s

ω(Ryan) = 2pif. = 2pi(300Hz) = 1883.96 rad/s

2. Incorporating angular velocities into the fav = (f1 + f2)/2 formula, we’re able to find the mean (average) angular frequency.

[ω(Lisa) + ω(Ryan)]/2 = [1231.50 + 1883.96]/2 = 1557.73 rad/s

3. We know 1Hz = [2pi rad]/s so 1557.73 rad/s = 247.92Hz

Question 2: A simultaneous sound is produced when two tuning forks are struck together. If one tuning fork has a frequency of 250Hz, and the other tuning fork has a frequency of 235 Hz, what is the beat frequency?

1. We’re looking for beat frequency, so we refer to the formula:fbeat = l f2 – f1 l

2. Given f1 = 250 Hz and f2 = 235 Hz, we simply plug in the numbers.

fbeat = l 235Hz – 250Hz l

fbeat = l -15Hz l

fbeat = 15Hz

3. So the beat frequency is 15 Hz

Important points about beats…

• A max. sound is heard, & amplitude is at its max. when 2 waves w/slightly different frequencies combine & their crests/troughs coincide

• A min. sound is heard, & amplitude is at zero when the waves go out of phase & the amplitude of the resultant waves decreases

• The rate at which amplitude varies is proportional to the frequency difference

Intensity of a Wave

• Defn. as avg rate per unit area

• Typically expressed in W/m^2

• At which energy is transported by the wave across a surface

^BASICALLY: intensity=avg power/unit sq.

OR intensity = power/area I α 1/r2

Intensity Level Equation

• Given by decibels:

= ( 10dB ) log ( I / lo )

the sound intensity level

I the sound intensity

Io = 10-12 W/m2 , the standard intensity

Question 4: The sound level 50 m away from a speaker system is =125 dB. What is the ratio of the intensity I2 of the speakers at the spot to the intensity I1 of a jackhammer operating at a sound level of =95 dB?

1. From the question, we know that we want the ratio I2 /I1.

2. How do we find the ratio? We need to find I2 and I1 separately, with the equation = (10dB )log(I/Io).

3. Now, plug in the values, remember Io= 10-12 W/m2:

2 = (10dB)log(I2 /Io) = 125 dB

1 = (10db)log(I1 /Io) = 95 dB

4. After solving for I2 and I1, you have the values for the ratio I2/I1.

I2/I1 = 10^[(125-95)/10] = 1000

Question 5: The drum’s sound source is increased 30dB in sound level. By what multiple is its intensity increased?

1. Since we’re dealing with sound level ratios again, we refer back to the equation = (10dB )log(I/Io).

2. We understand the difference between I1 and I2 is that the sound level INCREASED by 30dB in sound intensity level. This means we simply plug in 1 and 2 values with a difference of 30dB.

1 = (10dB)log(I1/Io) = 10 I1 = 10^(-11) W

2 = (10dB)log(I2/Io) = 40 I2 = 10^(-8) W

3. Use I2/I1 to find the ratio (the multiple).

I2/I1 = 10^(-8)/10^(-11) = 10^(3) So a multiple of 1000.