In the arrangement shown in the figure below, an object of mass m = 4.0 kg hangs

Question

In the arrangement shown in the figure below, an object of mass m = 4.0 kg hangs from a cord around a light pulley. The length of the cord between point P and the pulley is L = 2.0 m. (Ignore the mass of the vertical section of the cord.)

(a) When the vibrator is set to a frequency of 130 Hz, a standing wave with six loops is formed. What must be the linear mass density of the cord? kg/m

(b) How many loops (if any) will result if m is changed to 144 kg? (Enter 0 if no loops form.) loops

(c) How many loops (if any) will result if m is changed to 10 kg? (Enter 0 if no loops form.) loops

Explanation / Answer

(a)

frequency f = (6v/2L)

v = speed = sqrt(T/u)

T = tension in the string = mg = 4*9.8

u = linear density

130 = (6v/4)

speed v = 86.7 m/s

86.7 = sqrt(4*9.8/u)

linear density u = 5.21*10^-3 kg/m

+_____________________

(b)

v = sqrt(144*9.8/(5.21*10^-3)) = 520.45 m/s

f = nv/2L

130 = (n*520.45)/4

n =0.99

0 loops

_____________________________

(c)

v = sqrt(10*9.8/(5.21*10^-3)) = 137.15 m/s

f = nv/2L

130 = (n*137.15)/4

n = 3

3 loops

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