A tuning fork and an air column whose tempera | vedclass

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(D) Let the frequency of the tuning fork be $f$ and the frequency of the air column at temperature $T$ be $n_T$. Since the speed of sound $v \propto \

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$50$

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(D) Let the frequency of the tuning fork be $f$ and the frequency of the air column at temperature $T$ be $n_T$. Since the speed of sound $v \propto \sqrt{T}$,the frequency $n_T \propto \sqrt{T}$.At $51^{\circ} C$ $(324 \, K)$,the beat frequency is $4$,so $n_{324} = f \pm 4$.At $16^{\circ} C$ $(289 \, K)$,the beat frequency is $1$,so $n_{289} = f \pm 1$.Since the number of beats decreases as temperature decreases,the frequency of the air column must be higher than the tuning fork frequency $(n_T > f)$.Thus,$n_{324} = f + 4$ and $n_{289} = f + 1$.Taking the ratio: $\frac{n_{324}}{n_{289}} = \sqrt{\frac{324}{289}} = \frac{18}{17}$.Substituting the expressions: $\frac{f+4}{f+1} = \frac{18}{17}$.$17(f+4) = 18(f+1) \implies 17f + 68 = 18f + 18$.$f = 68 - 18 = 50 \, Hz$.

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