In the arrangement shown in Figure P14.40, an object of mass, m = 3.0 kg, hangs

Question

In the arrangement shown in Figure P14.40, an object of mass, m = 3.0 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.

(a) When the vibrator is set to a frequency of 170 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 108 kg?
______ loops

(c) How many loops (if any) will result if m is changed to 4.32 kg?
_______loops

Explanation / Answer

n = 6 so f6 = 170 = 6*v/2L but v = sqrt(T/)

And T = m*g

so 170 = 6*sqrt(T/)/2L

So = (6/170)^2*T/4L^2 = (6/170)^2*m*g/(3L^2) = (6/170)^2*3.0*9.8/(3*2^2) = 3.051x10^-3kg/m

b) Now f= 130 = n*sqrt(T/)/(2L)

So n = 170*2*L/sqrt(m*g/) = 170*2*2.0m/sqrt(108.0*9.8/3.051x10^-3) = 1.154

c) Now f= 170 = n*sqrt(T/)/(2L)

So n = 170*2*L/sqrt(m*g/) = 170*2*2.0m/sqrt(4.32*9.8/3.051x10^-3) = 5.772

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