Answer to Question #163184 in Optics for Vanel

Question

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?

Answer 1

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

a)

v = $\lambda$f

L = 0.5n$\lambda$

$\lambda$ = 2L / n

v = (T / M)^1/2

M =$\frac{T}{v^2} = \frac{T}{\lambda^2f^2} =$ $\frac{Tn^2}{4L^2f^2} = 0.0049 \frac{kg}{m}$

b) m = 45 kg

45 / 5 = 9

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

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

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