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What will be the "beat frequency" if middle C(262 Hz) and C#(277Hz) are played together? ... Hz ...

Question:

What will be the "beat frequency" if middle C(262 Hz) and C#(277Hz) are played together? _____ Hz

1. Will this be audible?

a. No, this can not be heard by the human ear.

b. Yes, this can be heard by the human ear.

2. What if each is played two octaves lower (each frequency reduced by a factor of 4)?

a. Yes, this can be heard by the human ear.

b. No, this can not be heard by the human ear.

Beat Formation

When two waves of comparable frequency superpose with each other, they form beat whose frequency is given by

{eq}\begin{align} f_{beat} = f_1 - f_2 \end{align} {/eq}

Where {eq}\ f_1 \ \& \ f_2 {/eq} are the frequencies of wave one and wave two.

Answer and Explanation:

Data Given

  • Frequency of middle C {eq}f_2 = 262 \ \rm Hz {/eq}
  • Frequency of middle C# {eq}f_1 = 277 \ \rm Hz {/eq}

Part A) the beat frequency is given by

{eq}\begin{align} f_{beat} = f_1 - f_2 \end{align} {/eq}

{eq}\begin{align} f_{beat} = 277 \ \rm Hz - 262 \ \rm Hz \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{ f_{beat} = 15 \ \rm Hz }} \end{align} {/eq}

As the beat frequency is 15 Hz is less then the 20 Hz so the beat cannot be heard by the human ear.

Part B) Again

{eq}\begin{align} f'_{beat} =\frac{1}{4} f_1 - \frac{1}{4} f_2 \end{align} {/eq}

{eq}\begin{align} f'_{beat} = \frac{277 \ \rm Hz}{4} - \frac{262 \ \rm Hz}{4} \end{align} {/eq}

{eq}\begin{align} f'_{beat} = 69.25 \ \rm Hz - 65.5 \ \rm Hz \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{ f'_{beat} = 3.75 \ \rm Hz }} \end{align} {/eq}

As the beat frequency is 3.75 Hz is less then the 20 Hz so the beat can not be heard by the human ear.


Learn more about this topic:

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Constructive and Destructive Interference

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16
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