A spider has a mass 0.70 g and is hanging from rest on a 20-cm-long thread of 0.

Question

A spider has a mass 0.70 g and is hanging from rest on a 20-cm-long thread of 0.0020-mm-diameter silk. Spider silks density is 1290 kg/m3. (a) If the small child holding the other end of the silk moves their end of the silk, how long will it take for the spider to feel it? (b) If the string is whirled in a circle at a rate of 2 revolutions per second, what is the minimum and maxi-mum time for vibrational information from one side of the string to reach the other?

Please show/explain all steps so I can understand!!

Explanation / Answer

cross sectional area = pi*(1*10^-6)^2 = 3.141*10^-12 m^2

mass per unit length mu = density*area

= 1290*3.141*10^-12

= 4.0526*10^-9 Kg/m

Wave velocity V = sqrt(T/mu) = sqrt(m*g/ mu)

= sqrt(0.0007*9.81 / 4.0526*10^-9)

= 1301.7 m/sec

Time taken = distance/speed = 0.2/1301.7 = 1.536*10^-4 sec = 0.153 ms

centipetal force = m*r*omega^2

= 0.0007*0.2*(2*pi*2)^2

= 0.0221 N

maximum tension = 0.0221 + mg = 0.0221 + (0.0007*9.81)

= 0.0289 N

V = sqrt(T/mu) =sqrt(0.0289/4.0526*10^-9) = 2673.89 m/sec

minimum time = 0.2/2673.89 = 0.0747 ms

minimum tension = 0.0221 - mg = 0.01523 N

V = sqrt(T/mu) =sqrt(0.01523/4.0526*10^-9) = 1938.76 m/sec

maximum time = 0.2/1938.76 = 0.103 ms

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