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JEE Main 2006
Exam
JEE Main
Year
2006
Subject
Physics
Chapter
Waves and Sound

JEE Main 2006 · Physics · Waves and Sound

A string is stretched between fixed points separated by 75 ~cm . It...

A string is stretched between fixed points separated by 75 cm75 \mathrm{~cm}. It is observed to have resonant frequencies of 420 Hz420 \mathrm{~Hz} and 315 Hz315 \mathrm{~Hz}. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Options

  1. A10.5 Hz10.5 \mathrm{~Hz}
  2. B105 Hz105 \mathrm{~Hz}
  3. C1.05 Hz1.05 \mathrm{~Hz}
  4. D1050 Hz1050 \mathrm{~Hz}

Correct answer

B. 105 Hz105 \mathrm{~Hz}

Step-by-step solution

n2ℓ(v)=315,(n+1)2ℓv=420\frac{\mathrm{n}}{2 \ell}(\mathrm{v})=315, \frac{(\mathrm{n}+1)}{2 \ell} \mathrm{v}=420 Solving v2ℓ=105\frac{\mathrm{v}}{2 \ell}=105
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