In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4.0 cm. If the speed of sound in air is $328 m s^{- 1}$ , what is the frequency of the source ?

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Solution

Here given distance between two nodes is=4.0 cm

$\Rightarrow λ = 2 \times 4.0 = 8 c m$

We know that, $v = n λ$

$\Rightarrow n = \frac{328}{8 \times 10^{- 2}} = 4.1 K H z$

V=340 m/s

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In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4.0 cm. If the speed of sound in air is 328ms−1, what is the frequency of the source ?

Here given distance between two nodes is=4.0 cm ⇒λ=2×4.0=8cm We know that, v=nλ ⇒n=3288×10−2=4.1KHz V=340 m/s

In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4.0 cm. If the speed of sound in air is $328 m s^{- 1}$ , what is the frequency of the source ?

Here given distance between two nodes is=4.0 cm

$\Rightarrow λ = 2 \times 4.0 = 8 c m$

We know that, $v = n λ$

$\Rightarrow n = \frac{328}{8 \times 10^{- 2}} = 4.1 K H z$

V=340 m/s