A tuning fork of known frequency 256,Hz produ | vedclass

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(D) Let the initial frequency of the guitar string be $f$. The beat frequency is given by $|f - 256| = 5$. This implies $f = 256 \pm 5$,so $f = 261\,H

Vedclass · Accepted Answer

$251$

Vedclass · Answer

(D) Let the initial frequency of the guitar string be $f$. The beat frequency is given by $|f - 256| = 5$. This implies $f = 256 \pm 5$,so $f = 261\,Hz$ or $f = 251\,Hz$.When the tension in the string is increased,the frequency $f$ of the string increases $(f \propto \sqrt{T})$.Case $1$: If $f = 261\,Hz$,increasing the tension will make $f > 261\,Hz$,so the beat frequency $|f - 256|$ will increase beyond $5$.Case $2$: If $f = 251\,Hz$,increasing the tension will make $f$ approach $256\,Hz$. The new beat frequency is $|f' - 256| = 2$. Since $f' > 251$,$f'$ must be $254\,Hz$ (as $256 - 254 = 2$).Thus,the initial frequency was $251\,Hz$.

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