One end of a horizontal string of linear density 4.24 times 104kg/m is attached

Question

One end of a horizontal string of linear density 4.24 times 104kg/m is attached to a small-amplitude mechanical 51.7Hz vibrator. The string passes over a pulley, a distance L = 1.58m away, and weights are hung from this end, as seen in the figure below. What mass must be hung from this end of the string to produce one loop of a standing wave? Assume the string at the vibrator is a node, which is nearly true. What mass must be hung from this end of the string to produce two loops of a standing wave?

Explanation / Answer

f = nv/ 2l

n = 2

v = T/

T = mg

51.7 = 2 x (m x 9.8/4.24 x 10-4)/ 2 x 1.58

m = 0.28 kg

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