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

Question

Eshan · Answer

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.

Hence $6\times \dfrac{\lambda}{2}=1.5m\implies \lambda=0.5m$
Distance between node and antinode is half the length of each loop

= $\dfrac{1}{2}\times\dfrac{\lambda}{2}=\dfrac{\lambda}{4}=0.125m$

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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

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

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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

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

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