In a stationary wave the distance between consecutive antinodes is $25 c m$ . If the wave velocity is $300 m s^{- 1}$ then the frequency of wave will be

150 H z

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300 H z

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600 H z

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750 H z

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Solution

The correct option is C $600 H z$
Distance between two consecutive antinodes is half the wavelength.

Frequency, $f = \frac{V}{λ}$

$λ = 50 c m = 0.5 m$

$f = \frac{300}{0.5} = 600 H z$

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In a stationary wave the distance between consecutive antinodes is 25 cm. If the wave velocity is 300 ms−1 then the frequency of wave will be

The correct option is C 600 HzDistance between two consecutive antinodes is half the wavelength.Frequency, f=Vλλ=50 cm=0.5 mf=3000.5=600 Hz

In a stationary wave the distance between consecutive antinodes is $25 c m$ . If the wave velocity is $300 m s^{- 1}$ then the frequency of wave will be

The correct option is C $600 H z$
Distance between two consecutive antinodes is half the wavelength.

Frequency, $f = \frac{V}{λ}$

$λ = 50 c m = 0.5 m$

$f = \frac{300}{0.5} = 600 H z$