Question

(b) The particle displacement y of air molecules due to a sound wave is given by y 4m and w = = 0.008 cos wt sin kz. Where k - m 50m rad/s. calculate i. The distance between two consecutive nodes ii. The amplitude after 0.56s​

Answer 1

The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.

What's the distance between consecutive nodes of the displacement of air molecules?

• Wavelength is the distance between two consecutive nodes or toughs or crests or anti-nodes.
• So, distance between consecutive nodes = wavelength = 2π÷k

= 2π/(4π÷m)

= m/2

What's the amplitude after 0.56s of the displacement of air molecules?

Displacement after 0.56 s = 0.008×cos(50π×0.56s)

=1.75×10^(-4) m

Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.

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Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.

Calculate:

I) the distance between 2 consecutive nodes

ii) the amplitude after 0.565s

Learn more about the wavelength here:

brainly.com/question/10750459

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