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(A) Given: Speed of student $v = 5.4 \,km/h = 5.4 	imes \frac{5}{18} \,m/s = 1.5 \,m/s$.Distance of pipe opening from the path $D = 8 \,m$.Diameter o

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$6-12$

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(A) Given: Speed of student $v = 5.4 \,km/h = 5.4 	imes \frac{5}{18} \,m/s = 1.5 \,m/s$.Distance of pipe opening from the path $D = 8 \,m$.Diameter of the pipe $d = 0.45 \,m$.Frequency of sound $f = 1280 \,Hz$.Speed of sound $v_s = 320 \,m/s$.First,calculate the wavelength of the sound: $\lambda = \frac{v_s}{f} = \frac{320}{1280} = 0.25 \,m$.The diffraction of sound waves at the circular opening of the pipe follows the condition for the first minimum: $\sin 	heta = 1.22 \frac{\lambda}{d}$.For small angles,$\sin 	heta \approx 	an 	heta = \frac{y}{D}$,where $y$ is the distance from the center of the central maximum to the first minimum.$y = 1.22 \frac{\lambda}{d} D = 1.22 	imes \frac{0.25}{0.45} 	imes 8 = 1.22 	imes \frac{1}{1.8} 	imes 8 \approx 5.42 \,m$.The width of the central maximum is $2y = 2 	imes 5.42 = 10.84 \,m$.The time $T$ for which the sound is heard is the time taken to cover this width: $T = \frac{2y}{v} = \frac{10.84}{1.5} \approx 7.23 \,s$.This value lies in the range of $6-12$ seconds.

Similar Questions

In Fraunhofer diffraction pattern, slit width is $0.2 \, mm$ and screen is at $2 \, m$ away from the lens. If wavelength of light used is $5000 \, \mathring{A}$, then the distance between the first minimum on either side of the central maximum is (assume $\theta$ is small and measured in radians).

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