MCQ
A tuning fork of frequency $280\,\, Hz$ produces $10$ beats per sec when sounded with a vibrating sonometer string. When the tension in the string increases slightly, it produces $11$ beats per sec. The original frequency of the vibrating sonometer string is ... $Hz$
  • A
    $269$
  • B
    $291$
  • C
    $270 $
  • $290$

Answer

Correct option: D.
$290$
d
The frequency of tuning fork is $\nu_{T}=280 \mathrm{Hz}$

The frequency of sonometer string is $\nu_{s}$

The frequency of beats is

$\left|\nu_{T}-\nu_{s}\right|=10$

Hence, the frequency of sonometer string is either $10 \mathrm{Hz}$ more i.e. $290 \mathrm{Hz}$ or $10 \mathrm{Hz}$ lessi.e. $270 \mathrm{Hz}$ because increase in this frequency will reduce the frequency of beats.

But when tension is applied to the string its frequency increases and number of beats also increases. Hence, frequency of string cannot be $290 \mathrm{Hz}$.

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