Question

A 107 kg mass suspended from a wire whose unstretched length is 4.05 m is found to stretch the wire by 6.40 mm . The wire has a uniform cross-sectional area of 0.130 cm².

a) If the load is pulled down a small additional distance and released, find the frequency at which it vibrates (in Hz).

b) Compute Young's modulus for the wire.

Answer 1

(a) Expansion in the wire, x = 6.40 mm = 0.0064 m

Force on the wire, F = m*g = 107*9.8 = 1048.6 N

Suppose the spring constant of the wire = k

Use the expression -

F = k*x

=> k = F/x = 1048.6 / 0.0064 = 163843.75 N/m

Frequency of oscillation, f = (1/2pi)*sqrt[k/m] = (1/2pi)*sqrt[163843.75 / 107] = 6.23 Hz

(b) Cross-sectional area of the wire, A = 0.130 cm^2 = 0.130 x 10^-4 m^2

Stress = F/A = 1048.6 / (0.130 x 10^-4) = 8066 x 10^4 N/m^2

Strain = x / L = 0.0064 / 4.05 = 0.00158

So, Young's modulus of wire, Y = Stress / Strain = (8066 x 10^4) / 0.00158 = 5.105 x 10^10 N/m^2