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 180 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.)

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

Explanation / Answer

a] wavelength lambda = L/3 = 2/3 m

speed = lambda*f = 2/3*180 = 120 m/s

v = sqrt(T/u)

120^2 = T/u

u = T/120^2 = [4*9.8]/120^2 = 0.00272 kg/m

b] v = sqrt(144*9.8/0.00272) = 720 m/s

lambda = v/f = 720/180 = 4m = 2L

loop = 2*L/2L

= 1 loop answer

c] v = sqrt(10*9.8/0.00272) = 189.8 m/s

lambda = v/f = 189.8/180 =1.0544 m

loop = 2*L/lambda = 4/1.0544 =

= 3.8 loop

on rounding off, 4 loop. answer

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