In the arrangement shown in the figure below, an object of mass m = 2.00 kg hang

Question

In the arrangement shown in the figure below, an object of mass m = 2.00 kg hangs from a cord around a light pulley. The length of the cord between point P and the pulley is L = 2.00 m. (Ignore the mass of the vertical section of the cord.)

(a) When the vibrator is set to a frequency of 154 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 4.50 kg? (Enter 0 if no loops form.)

(c) How many loops (if any) will result if m is changed to 10.0 kg? (Enter 0 if no loops form.)

Explanation / Answer

  Here we have n = 6 so f6 = 154 = 6*v/2L but v = sqrt(T/)

And T = m*g

so 154 = 6*sqrt(T/)/2L

So = (6/154)^2*T/4L^2 = (6/154)^2*m*g/(4L^2) = (6/154)^2*2.0*9.8/(4*2^2) = 1.86x10^-3kg/m

b) Now f= 154 = n*sqrt(T/)/(2L)

So n = 154*2*L/sqrt(m*g/) = 154*2*2.0m/sqrt(4.5*9.8/1.86x10^-3) = 4

c) Now f= 154 = n*sqrt(T/)/(2L)

So n = 154*2*L/sqrt(m*g/) = 154*2*2.0m/sqrt(10.0*9.8/1.86x10^-3) = 2.68 = 3

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