Stationary waves of frequency 200 Hz are formed in air. If the velocity of the wave is 360 m/s, the shortest distance between two antinodes is

Question

Stationary waves of frequency 200 Hz are formed in air. If the velocity of the wave is 360 m/s, the shortest distance between two antinodes is (A) 1.8 m (B) 3.6 m (C) 0.9 m (D) 0.45 m

Tardigrade Admin · Accepted Answer

Shortest distance between two antinodes is equal to the half of wavelength. Here, v = 360 m s-1 , upsilon = 200 Hz ∴ λ = (v/ upsilon) = (360/200) = 1.8 m Hence (λ /2) = (1.8/2) = 0.9 m

Q. Stationary waves of frequency 200 Hz are formed in air. If the velocity of the wave is 360 m/s, the shortest distance between two antinodes is

1.8 m

3.6 m

0.9 m

0.45 m

Shortest distance between two antinodes is equal to the half of wavelength.
Here, $v = 360 m s^{- 1}, υ = 200 Hz$
$∴ λ = \frac{v}{υ} = \frac{360}{200} = 1.8 m$
Hence $\frac{λ}{2} = \frac{1.8}{2} = 0.9 m$

Q. Stationary waves of frequency 200 Hz are formed in air. If the velocity of the wave is 360 m/s, the shortest distance between two antinodes is

1.8 m

3.6 m

0.9 m

0.45 m

Shortest distance between two antinodes is equal to the half of wavelength. Here, v=360ms−1,υ=200Hz ∴λ=υv​=200360​=1.8m Hence 2λ​=21.8​=0.9m

Shortest distance between two antinodes is equal to the half of wavelength.
Here, $v = 360 m s^{- 1}, υ = 200 Hz$
$∴ λ = \frac{v}{υ} = \frac{360}{200} = 1.8 m$
Hence $\frac{λ}{2} = \frac{1.8}{2} = 0.9 m$