A tuning fork A of frequency 250 ,Hz and anot | vedclass

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Question

(A) Given: Frequency of tuning fork $A$ $(f_A)$ = $250 \,Hz$. Frequency of tuning fork $B$ $(f_B)$ = $x \,Hz$.Initially, the beat frequency is $5 \,Hz

Vedclass · Accepted Answer

$255$

Vedclass · Answer

(A) Given: Frequency of tuning fork $A$ $(f_A)$ = $250 \,Hz$. Frequency of tuning fork $B$ $(f_B)$ = $x \,Hz$.Initially, the beat frequency is $5 \,Hz$, so $|f_A - f_B| = 5$.This implies $250 - x = 5$ or $x - 250 = 5$, giving $x = 245 \,Hz$ or $x = 255 \,Hz$.When tuning fork $B$ is waxed, its frequency decreases $(f_B' After waxing, the new beat frequency is $3 \,Hz$.If $x = 255 \,Hz$, then $f_B$ decreases towards $250 \,Hz$, so the beat frequency would increase from $5$ to a higher value or pass through $0$ and then increase. This does not match the observation of $3 \,Hz$.If $x = 245 \,Hz$, then $f_B$ decreases further away from $250 \,Hz$ (e.g., to $244 \,Hz$), which would increase the beat frequency to $6 \,Hz$. However, if we consider the frequency of $B$ was $245 \,Hz$ and it was slightly loaded, the beat frequency decreases from $5$ to $3$ only if the frequency of $B$ was originally $255 \,Hz$ and it decreased to $253 \,Hz$ ($255 - 253 = 2$ - incorrect) or if $x = 245 \,Hz$ and the beat frequency changed. Let's re-evaluate: If $x = 255 \,Hz$, loading $B$ makes $f_B$ approach $250 \,Hz$, decreasing the beat frequency. Thus, $x = 255 \,Hz$ is the correct value.

$A$ tuning fork $A$ of frequency $250 \,Hz$ and another tuning fork $B$ of frequency $x$ produce $5$ beats per second when vibrated together. If the fork $B$ is waxed and vibrated together with $A$, then $3$ beats per second are produced. Then $x=$ (in $\,Hz$)

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