One end of a horizontal string is attached to a small-amplitude mechanical 60.0

Question

One end of a horizontal string is attached to a small-amplitude mechanical 60.0 Hz oscillator. The string's mass per unit length is 3.9 x 10^{-4} kg/m. The string passes over a pulley, a distance L = 1.50 m away, and weights are hung from this end. Assume the string at the oscillator is a node, which is nearly true.

One end of a horizontal string is attached to a small-amplitude mechanical 60.0 Hz oscillator. The string's mass per unit length is 3.9 x 10^{-4} kg/m. The string passes over a pulley, a distance L = 1.50 m away, and weights are hung from this end. Assume the string at the oscillator is a node, which is nearly true. What mass m must be hung from this end of the string to produce one loop of a standing wave? What mass m must be hung from this end of the string to produce two loops of a standing wave? What mass m must be hung from this end of the string to produce five loops of a standing wave?

Explanation / Answer

A.

for one loop

lamda/2 = L

hence lambda = 3

v = f*lambda

v= 60*3= 180 m/s

v = sqrt (T/3.19*10^-4)

T= mg

180 = sqrt (T/3.19*10^-4)

180*180 = m*9.8/(3.19*10^-4)

B.

for 2 loops

lamda = L

hence lambda = 1.5

v = f*lambda

v= 60*1.5= 60 m/s

v = sqrt (T/3.19*10^-4)

T= mg

60 = sqrt (T/3.19*10^-4)

60*60 = m*9.8/(3.19*10^-4)

C.

for 5 loops

lambda = 2L/5

hence lambda = 0.6

v = f*lambda

v= 60*0.6= 36 m/s

v = sqrt (T/3.19*10^-4)

T= mg

36 = sqrt (T/3.19*10^-4)

36*36 = m*9.8/(3.19*10^-4)

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