A 13 -kg object hangs in equilibrium from a string of total length L = 5.0 m and

Question

A 13-kg object hangs in equilibrium from a string of total length L = 5.0 m and linear mass density µ = 0.0010 kg/m. The string is wrapped around two light, frictionless pulleys that are separated by the distance d = 2.0 m (Fig. a).

A 13-kg object hangs in equilibrium from a string of total length L = 5.0 m and linear mass density mu = 0.0010 kg/m. The string is wrapped around two light, frictionless pulleys that are separated by the distance d = 2.0 m (Fig. a). (a) Determine the tension in the string. N (b) At what frequency must the string between the pulleys vibrate in order to form the standing-wave pattern shown in Figure (b)? Hz

Explanation / Answer

L = 5m
d = 2m
µ = 0.0010 kg/m
m= 13kg
x1+x2=3m
x1=x2=1.5m

http://i289.photobucket.com/albums/ll238/ninditsu/trigpic-1.png
by trig based on length of string (look at picture)
=arccos[(d/2)/x1]

http://i289.photobucket.com/albums/ll238/ninditsu/forces.png
create a free body diagram for the hanging mass (look at picture)
T1=T2
so forces in y direction because mass is hanging in equilibrium 2*T1sin=mg
in x direction the only forces acting are the 2 tensions and they cancel
solve for T1. T1 =0.5*mg/sin?

substitute givens and calculated tension.
it is 3rd harmonic (3 loops)
f3=frequency=1.5 * [sqrt(T/µ)] / L

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