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NEET 2010
Exam
NEET
Year
2010
Subject
Physics
Chapter
Waves and Sound

NEET 2010 · Physics · Waves and Sound

A tuning fork of frequency 512 ~Hz makes 4 beats/seconds with the...

A tuning fork of frequency 512 Hz512 \mathrm{~Hz} makes 4 beats/seconds with the vibrating string of a piano. The beat frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

Options

  1. A510 Hz510 \mathrm{~Hz}
  2. B514 Hz514 \mathrm{~Hz}
  3. C516 Hz516 \mathrm{~Hz}
  4. D508 Hz508 \mathrm{~Hz}

Correct answer

D. 508 Hz508 \mathrm{~Hz}

Step-by-step solution

Suppose np=n_p= frequency of piano = ? (np∝T)nf=\left(n_p \propto \sqrt{T}\right) \mathrm{n}_{\mathrm{f}}= frequency of tuning fork =512 Hzx=512 \mathrm{~Hz} \mathrm{x} = Beat frequency =4=4 beats /s/ \mathrm{s}, which is decreasing (4→2)(4 \rightarrow 2) after changing the tension of piano wire. Also, tension of piano wire is increasing so np↑n_p \uparrow Hence, np↑−nf=x↓⟶\mathrm{n}_{\mathrm{p}} \uparrow-\mathrm{n}_{\mathrm{f}}=\mathrm{x} \downarrow \longrightarrow wrong nf−np↑=x↓⟶\mathrm{n}_{\mathrm{f}}-\mathrm{n}_{\mathrm{p}} \uparrow=\mathrm{x} \downarrow \longrightarrow correjct ⇒np=nf−x=512−4=508 Hz\Rightarrow \mathrm{n}_{\mathrm{p}}=\mathrm{n}_{\mathrm{f}}-\mathrm{x}=512-4=508 \mathrm{~Hz}
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