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Two tunning forks have frequencies of 256 Hz. A bit of clay is placed on the prong of...

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how2becrazy | Student, Grade 11 | eNotes Newbie

Posted May 3, 2012 at 2:00 AM via web

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Two tunning forks have frequencies of 256 Hz. A bit of clay is placed on the prong of one tunning fork produces a beat frequency of 4 Hz when both are struck simultaneously. What is the frequency of the fork with the clay?

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mwmovr40 | College Teacher | (Level 1) Associate Educator

Posted May 8, 2012 at 12:29 AM (Answer #1)

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The frequency of a tuning fork, or any other vibrating object, is the number of times it completes a cycle per second.  Or: it is the amount of time for the pattern created by the vibration to repeat itself.  When two pure tones are created simultaneously the waves of sound created will overlap each other creating constructive and destructive interference.  Constructive interference is areas where wave forms add to each other; destructive interference is where the wave forms subtract one from the other.  The result of the interference is the creation of a new wave that has a wave structure created from the addition and subtraction of the combined waves.  This new wave has its own wavelength and frequency.

The wave produced in this manner is called a beat frequency.  Musicians have used this fact to help tune musical instruments for eons.  A trained ear hears the beat frequency produced by two instruments playing notes that are slightly "out of tune" (that is, they have differening frequencies).  By adjusting one, or both, instruments the musician eliminates the beat indicating that both instruments now have the same frequency.

The formula for beat frequency would be Fb = Abs(F2 - F1)

where Fb is the beat frequency and is given by the absolute value of the difference between the two frequencies which are interfering.

Absolute value equations of this nature have two solutions, so there are usually two possible equations for beats:

Fb = F2-F1  and Fb = -(F2-F1).   If we make F2 the 256Hz tuning fork and F1 the altered tuning fork frequency we get

F1 = Fb+F2  and F1 = F2-Fb

The solution for the altered frequency would give us two choices

F1 = 260 Hz or 252Hz.  We can assume that the clay will make the fork vibrate more slowly, so we can eliminate the first choice and settle on the new frequency of the altered fork being 252Hz.

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