Physics-
General
Easy

Question

Two strings X and Y of a sitar produce a beat frequency 4 H z. When the tension of the string Y is slightly increased the beat frequency is found to be 2 blank H z. If the frequency of X is 300 blank H z comma then the original frequency of Y was

  1. 296 Hz    
  2. 298 Hz    
  3. 302 Hz    
  4. 302 Hz    

The correct answer is: 296 Hz


    n subscript x end subscript equals 300 H z comma blank n subscript y end subscript equals blank ?
    x equals beat frequency equals 4 blank H z comma which is decreasing left parenthesis 4 rightwards arrow 2 right parenthesis after increasing the tension of the string y.
    Also tension of wire y increasing so n subscript y end subscript upwards arrow left parenthesis because n proportional to square root of T right parenthesis
    Hence n subscript x end subscript minus n subscript y end subscript upwards arrow equals x downwards arrow blank rightwards arrow C o r r e c t
    n subscript y end subscript upwards arrow negative n subscript x end subscript equals x downwards arrow blank rightwards arrow W r o n g
    rightwards double arrow n subscript y end subscript equals n subscript x end subscript minus x equals 300 minus 4 equals 296 H z

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