Question

stationary waves of frequency 200 are formed in air if velocity of the wave is 360 M per second the shortest distance between antinodes will be:

Answer 1

f=200
v=360m/s
d=?
f=1/t
200=1/t
t=1/200
d=vt=360×1/200=9/5=1.8

Answer 2

Given:

Frequency f = 200 Hz

velocity of wave = 360 m/s

Formula Used:

Shortest distance between anti nodes = $\frac {\lambda}{2}$

where, $\lambda$ is the wavelength of the wave.

$\lambda = v\times f$

Calculation:

Substitute the values:

$\lambda = 360 m/s \times 200 /s = 72000 m = 72 km\\\frac {\lambda}{2} = 36000 m =36 km$

Thus, the shortest distance between anti  nodes is 36 km.

Learn more about: nodes and anti nodes

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