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Question
I am confused about this problem. I know it involves beats but i do not know what to do. Any help would be greatly appreciated.

A violin player hears four beats per second when she compares her note to a 523 Hz tuning fork. She can match the frequency of the tuning fork by tightening her string slightly. What was her initial frequency?

Answer
The beat frequency is equal to the difference between two different frequencies. Since the thing fork was 523Hz, the beat frequency is 4Hz and the player needs to tighten the violin string (the frequency/pitch increases when the tension is increased) the frequency of the violin string must be 4 Hz LESS that the tuning fork or 519Hz.

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James J. Kovalcin

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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