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Question

Two identical stringed instruments have frequency 100 Hz. If tension in one of them is increased by 4% and they are sounded together then the number of beats in one second is

  1. 1    
  2. 8    
  3. 4    
  4. 2    

The correct answer is: 2


    Frequency of vibration in tight string
    n equals fraction numerator p over denominator 2 l end fraction square root of fraction numerator T over denominator m end fraction end root rightwards double arrow n proportional to square root of TÞfraction numerator capital delta n over denominator n end fraction equals fraction numerator capital delta T over denominator 2 T end fraction equals fraction numerator 1 over denominator 2 end fraction cross times left parenthesis 4 percent sign right parenthesis equals 2 percent sign
    Þ Number of beats = capital delta n equals fraction numerator 2 over denominator 100 end fraction cross times n equals fraction numerator 2 over denominator 100 end fraction cross times 100 equals 2

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