A tuning fork with frequency 800 ; Hz produce | vedclass

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(C) For a resonance column tube closed at one end,the resonance lengths $l_1, l_2, l_3, \dots$ are given by $l_n = (2n-1) \frac{\lambda}{4}$.The diffe

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$344$

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(C) For a resonance column tube closed at one end,the resonance lengths $l_1, l_2, l_3, \dots$ are given by $l_n = (2n-1) \frac{\lambda}{4}$.The difference between two successive resonance lengths is $\Delta l = l_{n+1} - l_n = \frac{\lambda}{2}$.Given $l_1 = 9.75 \; cm$ and $l_2 = 31.25 \; cm$,the difference is $\Delta l = 31.25 \; cm - 9.75 \; cm = 21.50 \; cm = 0.215 \; m$.Thus,$\frac{\lambda}{2} = 0.215 \; m$,which implies $\lambda = 0.43 \; m$.The speed of sound $v$ is given by $v = f \lambda$.Substituting the values $f = 800 \; Hz$ and $\lambda = 0.43 \; m$,we get $v = 800 	imes 0.43 = 344 \; m/s$.

$A$ tuning fork with frequency $800 \; Hz$ produces resonance in a resonance column tube with the upper end open and the lower end closed by a water surface. Successive resonances are observed at lengths $9.75 \; cm$,$31.25 \; cm$,and $52.75 \; cm$. The speed of sound in air is ...... $m/s$.

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