Question #163184

In the arrangement, an object of mass m =  5.00kg hangs from a cord around a light pulley.   the length of the cord between the point p and the pulley is L = 2.00 meters.

(a) when the vibrator is set to a  frequency of 150Hz,  a standing wave with 6 Loops is formed. What must be the linear mass density of the cord?

(b) How many Loops if any will result if m is changed to 45.0 kilograms?



1
Expert's answer
2021-02-22T11:05:22-0500

L = 2 m; n = 6; T = 49 N; f = 150 hz; m = 5 kg

a)

v = λ\lambdaf

L = 0.5nλ\lambda

λ\lambda = 2L / n

v = (T / M)1/2

M =Tv2=Tλ2f2=\frac{T}{v^2} = \frac{T}{\lambda^2f^2} = Tn24L2f2=0.0049kgm\frac{Tn^2}{4L^2f^2} = 0.0049 \frac{kg}{m}


b) m = 45 kg

45 / 5 = 9

n = 4L2f2MT=2LfMT\sqrt{\frac{4L^2f^2M}{T}} =2Lf\sqrt{\frac{M}{T}}

n = 69=2\frac{6}{\sqrt{9}} = 2


Answer: a) M = 0.0049kg; b) n = 2.


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