Q117 / 180 · NEET 2015 (Phase 2) · Physics

A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencier between these two. The lowest resonant frequency for this string is:

  • A
    205 Hz
  • B
    10.5 Hz
  • Correct answer
    C
    105 Hz
  • D
    155 Hz

Solution

Two consecutive resonant frequencies for a string fixed at both ends will be

nv2l  and n+1v2l

⇒    n+1v2l-nv2l=420-315

v2l=105 Hz

Which is the minimum resonant frequency.

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

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