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Stationery waves produced on a stretch string which is 1.5 m long and fixed at its ends if six loops are formed find them wavelength and the distance between the nodes and adjacent antinodes

Suniti Nandi , 8 Years ago
Grade 12
anser 1 Answers
Eshan
Dear student,

Each loop aquires a length of half the wavelength and the total of 6 loops must be equal to the total length of the string.

Hence6\times \dfrac{\lambda}{2}=1.5m\implies \lambda=0.5m
Distance between node and antinode is half the length of each loop

=\dfrac{1}{2}\times\dfrac{\lambda}{2}=\dfrac{\lambda}{4}=0.125m
Last Activity: 7 Years ago
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