The frequency # (n)# of bouncing during hanging of a mass of #m# at the the of the rope along its length is related with its force constant #k# as follows
# n=1/(2pi)sqrt(k/m)#
Given
#m=90kg#
#k=1.4xx10^4N"/"m#
#n=?#
a) # n=1/(2pi)sqrt(k/m)#
#=> n=1/(2pi)sqrt((1.4xx10^4)/90)~~1.99Hz#
b) Let this rope stretches #x# m to break the climber’s fall, if he free-falls 2.00 m before the rope runs out of slack. Considering the conservation of energy we can say that the PE stored in the stretched nylon rope will be equal to loss of PE due to decrease in position of height #(x+2)# m.
Hence
#1/2kx^2=mxxgxx(x+2)#
#=>1/2xx1.4xx10^4xxx^2=90xx9.8xx(x+2)#
#=>7000x^2-882x=1764=0#
#=>x=(882+sqrt(882^2 +4xx7000xx1764))/14000~~0.569m#
(neglecting negative value of x)
c} The case where twice this length of nylon rope is used may be treated as two parallel rope of same length. In this case the equivalent force constant will be just twice i.e. #k=2.8xx10^4 N"/"m#, Substituting this value of k we can easily repeat the same calculation using above equations.
