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Question

In a stationary wave the distance between consecutive antinodes is $$25\space cm$$. If the wave velocity is $$300\space ms^{-1}$$ then the frequency of wave will be


A
150 Hz
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B
300 Hz
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C
600 Hz
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D
750 Hz
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Solution

The correct option is C $$600\space Hz$$
Distance between two consecutive antinodes is half the wavelength.

Frequency, $$f = \dfrac{V}{ \lambda} $$

$$\lambda = 50\ cm = 0.5\ m$$

$$f = \dfrac{300}{0.5} = 600\ Hz$$

Physics

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