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1) How do the positions of the nodes and the number of nodes you measured for each harmonic compare with what you expected? Explain.

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Answer #1

Amplitude of the stationary wave is given by

2 A = 2a[cos( )|

Now note that at nodes amplitude is zero

2a cos

cos(-) = 0 COS

herefore rac{2pi x}{lambda } = rac{pi}{2} , rac{3pi }{2} , rac{5pi }{2} ...

·x = (2p-1 ) 스 (1) 는 (Answer)

where p = 1, 2, 3...

for

x = rac{lambda }{4},rac{3lambda }{4}, rac{5lambda }{4}...

nodes are produced. Equation (1) reveals how position of nodes and number of nodes can be compared.

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