question_answer

Two progressive sound waves each of frequency \[\mathbf{170}\text{ }\mathbf{Hz}\] and travelling in opposite directions in air superpose to produce stationary waves. The speed of sound in air is\[\mathbf{340 m}{{\mathbf{s}}^{\mathbf{-1}}}\]. What is the separation between (i) two successive nodes, (ii) two successive antinodes and (iii) a node and its nearest antinode?

Answer:

                Wavelength of sound, \[\lambda =\frac{\upsilon }{v}=\frac{340}{170}=2.0m\] (i)Distance between two successive nodes \[\text{=  }\!\!\lambda\!\!\text{ /2 =}\mathbf{1}\mathbf{.0m}\] (ii) Distance between two successive antinodes \[=\lambda /2=\mathbf{1}\mathbf{.0m}\] (iii) Distance between a node and its nearest antinode \[=\lambda /4=\mathbf{0}\mathbf{.5m}.\]