A tuning fork of frequency $512\, Hz$ makes $4$ beats per second with the vibrating string of a piano. The beat frequency decreases to $2$ beats per sec when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was .... $Hz$
AIPMT 2010, Medium
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Let the frequencies of tuning fork and piano string be $v_{1}$ and $v_{2}$ respectively.
$\therefore v_{2}=v_{1} \pm 4=512 \mathrm{Hz} \pm 4=516 \mathrm{Hz}$ or $508 \mathrm{Hz}$
Increase in the tension of a piano string increases its frequency.
If $v_{2}=516 \mathrm{Hz}$, further increase in $v_{2},$ resulted in an increase in the beat frequency. But this is not given in the question.
If $v_{2}=508 \mathrm{Hz}$, further increase in $v_{2}$ resulted in decrease in the beat frequency. This is given in the question. When the beat frequency decreases to $2$ beats per second. Therefore, the frequency of the piano string before increasing the tension was $508\,\mathrm{Hz}$.
$\mathop {512\,Hz}\limits_{({v_1})} \,\,\xrightarrow{{ + \,\,4\,\,Hz}}\,\mathop {516\,Hz}\limits_{({v_2})} $
$\mathop {512\,Hz}\limits_{({v_1})} \,\,\xrightarrow{{ - \,\,4\,\,Hz}}\,\mathop {508\,Hz}\limits_{({v_2})} $
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