Q118 / 200 · NEET 2010 (Screening) · Physics

A tuning fork of frequency $512 \mathrm{~Hz}$ 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

  • A
    $510 \mathrm{~Hz}$
  • B
    $514 \mathrm{~Hz}$
  • C
    $516 \mathrm{~Hz}$
  • Correct answer
    D
    $508 \mathrm{~Hz}$

Solution

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

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

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