Question
In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4.0cm. If the speed of sound in air is 328m/s, what is the frequency of 73 the source?
✓
Answer
Here given distance between two nodes is = 4.0cm,$\Rightarrow\lambda=2\times4.0=8\text{cm}$
We know that $\text{V}=\text{n}\lambda$$\therefore \text{f}=\frac{\text{v}}{\lambda}=\frac{328}{8\times 10^{-2}}=4.1\text{KHz}$
$\eta=\frac{328}{8\times10^{-2}}=4.1\text{Hz}.$
We know that $\text{V}=\text{n}\lambda$$\therefore \text{f}=\frac{\text{v}}{\lambda}=\frac{328}{8\times 10^{-2}}=4.1\text{KHz}$
$\eta=\frac{328}{8\times10^{-2}}=4.1\text{Hz}.$
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