A spider has a mass 0.70g and is hanging from rest on a 20cm long thread of 0.00

Question

A spider has a mass 0.70g and is hanging from rest on a 20cm long thread of 0.0020 mm diameter silk. Spider silk's density is 1290kg/m^3.

(a) if a 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 vertical circle at a rate of 2 revolutions per second, what are the minimum and maximum time for vibrational information from one side of the string to reach the other?

Please show answers and solutions, including any equations used and how they are manipulated.

Explanation / Answer

Here,

mass of spider , M = 0.70 gm

M = 0.70 *10^-3 kg

a)

for time taken for wave to travel

time taken = distance/speed

time taken = L/sqrt(M*g/(m /L))

time taken = 0.20/sqrt(7 *10^-4 * 9.8/(1290 * pi * (0.002 *10^-3/2)^2))

time taken = 1.54 *10^-4 s

the time take for vibration to travel is 1.54 *10^-4 s

b)

for angular speed , w = 2 rev/s

w = 12.56 rad/s

angular acceleration ,a = w^2 * L

a = 12.56^2 * 0.20

a = 31.6 m/s^2

Now , for the maximum time

Tension in the string should be minimum

time taken = L/sqrt(M*(a - g)/(m /L))

time taken = 0.20/sqrt(7 *10^-4 * ( 31.6 - 9.8)/(1290 * pi * (0.002 *10^-3/2)^2))

time taken = 1.03 *10^-4 s

the maximum time taken is 1.03 *10^-4 s

for the minimum time

Tension must be maximum

time taken = L/sqrt(M*(a + g)/(m /L))

time taken = 0.20/sqrt(7 *10^-4 * ( 31.6 + 9.8)/(1290 * pi * (0.002 *10^-3/2)^2))

time taken = 7.47 *10^-5 s

the minimum time taken is 7.47 *10^-5 s

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