The string passes over a pulley, a distance l=1.52 m away, and weights are hung from this end, the figure (Figure 1) .
Part A
What mass m_1 must be hung from this end of the string to produce one loop of a standing wave? Assume the string at the oscillator is a node, which is nearly true.
Part B
What mass m_2 must be hung from this end of the string to produce two loops of a standing wave?
Part C
What mass m_3 must be hung from this end of the string to produce five loops of a standing wave?
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Verified answer
wave speed = v = lamda*f = (Tension/Mass Density)^(1/2) = (m*g/u)^(1/2)
solve for m and compute
f=240
u=5.0×10−4
For part A) lamda=l/2