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 160 Hz, a standing wave with six loops is formed. What must be the linear mass density of the cord? . Use the wavelength illustrated in the diagram and the other given information to calculate the wave speed and then from the wave speed find the linear mass density of the cord. kg/m (b) How many loops (if any) will result if m is changed to 72.0 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 = 130 = 6*v/2L )

but

) And T = m*g so 130 = 6*sqrt(T/?)/2L So ? = (6/130)^2*T/4L^2 = (6/130)^2*m*g/(4L^2) = (6/130)^2*4.0*9.8/(4*2^2) = 5.22x10^-3kg/m b) Now f= 130 = n*sqrt(T/?)/(2L) So n = 130*2*L/sqrt(m*g/?) = 130*2*2.0m/sqrt(9.0*9.8/5.22x10^-3) = 4 c) Now f= 130 = n*sqrt(T/?)/(2L) So n = 130*2*L/sqrt(m*g/?) = 130*2*2.0m/sqrt(36.0*9.8/5.22x10^-3) = 2

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