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(A) The frequency of the tuning fork is $f = 500 \,Hz$ and the velocity of sound is $v = 320 \,m/s$.The wavelength of the sound wave is $\lambda = \fr

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$3$

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(A) The frequency of the tuning fork is $f = 500 \,Hz$ and the velocity of sound is $v = 320 \,m/s$.The wavelength of the sound wave is $\lambda = \frac{v}{f} = \frac{320}{500} = 0.64 \,m = 64 \,cm$.For a tube closed at one end, resonance occurs when the length of the air column $L$ satisfies $L = \frac{(2n-1)\lambda}{4}$, where $n = 1, 2, 3, \dots$.The possible lengths for resonance are:For $n=1: L_1 = \frac{\lambda}{4} = \frac{64}{4} = 16 \,cm$.For $n=2: L_2 = \frac{3\lambda}{4} = 3 	imes 16 = 48 \,cm$.For $n=3: L_3 = \frac{5\lambda}{4} = 5 	imes 16 = 80 \,cm$.For $n=4: L_4 = \frac{7\lambda}{4} = 7 	imes 16 = 112 \,cm$.Since the total length of the tube is $1 \,m = 100 \,cm$, only lengths $16 \,cm, 48 \,cm,$ and $80 \,cm$ are possible.Therefore, the total number of resonances obtained is $3$.

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