Stationary waves of frequency 200 are formed in air. If velovity of the wave is 360 m/s the shortest distance between antinodes will be :

Stationary waves of frequency 200 are formed in air. If velovity of th

1.	Stationary waves of frequency 200 are formed in air. If velovity of the wave is 360 m/s the shortest distance between antinodes will be :
Answer» 1.8 m 3.6 m 2.7 m 0.9 m Solution :V = 200 HZ. V= 360 `ms^(-1)` `lambda= (u)/(v) = (360)/(200) = 1.8 `m distance bet. TWO conseutive ANTINODES `lambda// 2 = (18)/(2) = 0.9` m. Hence correct choice is (d).

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Stationary waves of frequency 200 are formed in air. If velovity of the wave is 360 m/s the shortest distance between antinodes will be :

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