In the arrangement shown in the figure below, an object of mass m = 6.0 kg hangs

Question

In the arrangement shown in the figure below, an object of mass m = 6.0 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?
kg/m

(b) How many loops (if any) will result if m is changed to 13.5 kg? (Enter 0 if no loops form.)
(c) How many loops (if any) will result if m is changed to 10 kg? (Enter 0 if no loops form.)

Explanation / Answer

part a:

six loops=3 wavelengths

==>3*wavelength=L= 2 m

==>wavelength=2/3 m

frequency=160 Hz

then speed of wave=wavelength*frequency=320/3 m/s

as we know, speed of a wave=sqrt(tension/mass per unit length)

tension in the string=m*g=6*9.8=58.8 N

then 320/3=sqrt(58.8/mass per unit length)

==>mass per unit length=58.8/(320/3)^2=0.0052 kg/m

b)

if m=13.5 kg,

tension=13.5*9.8=132.3 N

then speed of wave=sqrt(132.3/0.0052)=160 m/s

frequency=160 Hz

then wavelength=speed/frequnecy=1 m

hence number of loops=L/(0.5*wavelength)=2/0.5=4

part c:

if m=10 kg,

tension=10*9.8=98 N

then speed of wave=sqrt(98/0.0052)=137.706 m/s

frequency=160 Hz

then wavelength=speed/frequnecy=0.8607 m

hence number of loops=L/(0.5*wavelength)=9.2952=9 loops

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