Q89 / 160 · JIPMER 2011 · Physics

A tuning fork $A$ produces 4 beats/s with another tuning fork $B$ of frequency 320 Hz . On filing one of the prongs of $A, 4$ beats $/ \mathrm{s}$ are again heard when sounded with the same fork $B$. Then, the frequency of the fork $A$ before filing is

  • A
    328 Hz
  • B
    316 Hz
  • C
    324 Hz
  • Correct answer
    D
    320 Hz

Solution

There are 4 beats between $A$ and $B$, therefore the possible frequencies of $A$ are 316 or 324 that is $(320 \pm 4) \mathrm{Hz}$. When the prong of $A$ is filed, its frequency becomes greater than the original frequency. If we assume that original frequency of $A$ is 324 then on filing its frequency will be greater than 324 . The beats between $A$ and $B$ will be more than 4 . But it is given that the beats are again 4 , therefore, 324 is not possible. Therefore, required frequency must be 316 Hz . (This is true, because on filing the frequency may increase so as to give 4 beats with $B$ of frequency 320 Hz).

Hence, the correct answer is option: 'D'.

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