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PhysicsClass 11

In stationary waves, distance between a node and its nearest antinode is \(20\;cm\) . The phase difference between two particles having a separation of \(60\;cm\) will be     （ ） 
A. zero
B. \(\pi /2\)
C. \(\pi\)
D. \({3}\pi/{2}\)

Answer

D
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Solution

Given: distance between nearest node and anti node is \(20\;cm\) and seperation between two particles is, \(\triangle x=60\;cm\).
In a stationary wave there are two nodes and two antinodes.
\(\therefore\) Wavelength, \(\mathit{\lambda }=4\times 20=80\;cm\)
Phase difference, \(\triangle \mathit{\varPhi }=\dfrac{2\pi \triangle x}{\mathit{\lambda }}\)
In question \(\triangle x=\;60\;cm\)is given.
\(\triangle \mathit{\varPhi }=\dfrac{2\pi \times 60}{80}=\dfrac{3\pi }{2}\).
Hence the correct option is (\(D\)).
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