1. Tardigrade
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  4. When a tuning fork of frequency 341 is sounded with another tuning fork, six beats per second are heard. When he second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second timing fork is

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Q. When a tuning fork of frequency $341$ is sounded with another tuning fork, six beats per second are heard. When he second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second timing fork is

Waves

Solution:

$n_{A}=$ Known frequency $=341\, Hz \,, n_{B}=?$
$x=6$ bps, which is decreasing (i.e., $x \downarrow)$ after loading (from 6 to 1 bps) Unknown tuning fork is loaded so $n_{B} \downarrow$
Hence $n_{A}-n_{B} \downarrow=x \downarrow \ldots$ (i) Wrong
$n_{B} \downarrow-n_{A}=x \downarrow \ldots$ (ii) Correct
$\Rightarrow n_{B}=n_{A}+x=341+6=347\, Hz$