In the arrangement shown in the figure below, an object of mass m = 2.0 kg hangs

Question

In the arrangement shown in the figure below, an object of mass m = 2.0 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 140 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.)
__________loops

(c) How many loops (if any) will result if m is changed to 14 kg? (Enter 0 if no loops form.)
__________loops

Explanation / Answer

Here we have n = 6 so f6 = 140 = 6*v/2L but v = sqrt(T/?)

And T = m*g

so 140 = 6*sqrt(T/?)/2L

So ? = (6/140)^2*T/4L^2 = (6/140)^2*m*g/(4L^2) = (6/140)^2*2.0*9.8/(4*2^2) = 2.25x10^-3kg/m

b) Now f= 140 = n*sqrt(T/?)/(2L)

So n = 140*2*L/sqrt(m*g/?) = 140*2*2.0m/sqrt(4.5*9.8/2.25x10^-3) = 4

c) Now f= 140 = n*sqrt(T/?)/(2L)

So n = 140*2*L/sqrt(m*g/?) = 140*2*2.0m/sqrt(14.0*9.8/2.25x10^-3) = 2

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