Question

The number of beats produced per second by two tuning forks when sounded together is $4$.One of them has a frequency of $250Hz$ and if it is waxed, number of beats become $6$. The frequency of the other tuning fork is

• A. $254Hz$
• B. $252Hz$
• C. $248Hz$
• D. $246Hz$

Answer 1

Answer: (A)

$254Hz$

As per the problem, the frequency of beats is
$|{\nu }_1-{\nu }_2|=4$
where, ${\nu }_1$ and ${\nu }_2$ are the frequencies of given tuning forks.
Hence, the frequency of second tuning fork is either 4 Hz more i.e. 254 Hz or 4 Hz less i.e. 246 Hz than the first one.
Now, the first tuning fork is waxed this will reduce its frequency, hence the frequency of beats is now
${\nu }_1^{\prime }-{\nu }_2=6$
Since, decrease in frequency of first tuning fork increases the frequency of beats(by 2) hence, ${\nu }_1^{\prime }=248Hz$ and therefore for beat frequency 6 Hz, ${\nu }_2=254Hz$