Standing waves on a wire are described by y(x, t)=(A_SW sinkx sin omega t, with

Question

Standing waves on a wire are described by y(x, t)=(A_SW sinkx sin omega t, with A_ SW = 2.50mm , omega = 976rad/s , and k = 0.730pi ; rad/m . The left end of the wire is at x = 0. At what distances from the left end are the nodes of the standing wave. x_node} = 0.685+(1.37 m)n, n = 0, 1, 2, ... x_node} = (1.37 m)n, n = 0, 1, 2, ... x_node} = (0.685 m)n, n = 0, 1, 2, ... x_node} = (1.37cm)n, n = 0, 1, 2, ... At what distances from the left end are the antinodes of the standing wave? x_antinode} = 1.37 m + (1.37m)n, n = 0, 1, 2, ... x_ antinode} = 0.685 cm + (1.37 cm)n, n = 0, 1, 2, ... x_ antinode} = 0.685 m + (1.37 m)n, n = 0, 1, 2, ... x_antinode} = 0.685 m + (0.685 m)n, n = 0, 1, 2, ...

Explanation / Answer

x_node = 0.685+(1.37 m)n, n = 0, 1, 2, ...

x_node = (1.37 m)n, n = 0, 1, 2, ...

x_node = (0.685 m)n, n = 0, 1, 2, ...

x_node = (1.37cm)n, n = 0, 1, 2, ...

Solution:

= 2/k = 2/0.730 = 2.73

x_node = n (/2) = (1.37 m) n

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At what distances from the left end are the antinodes of the standing wave?

x_antinode} = 1.37 m + (1.37m)n, n = 0, 1, 2, ...

x_ antinode} = 0.685 cm + (1.37 cm)n, n = 0, 1, 2, ...

x_ antinode} = 0.685 m + (1.37 m)n, n = 0, 1, 2, ...

x_antinode} = 0.685 m + (0.685 m)n, n = 0, 1, 2, ...

Solution:

x_antinode = /4 + n /2 = 0.685 m + (1.37 m) n

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