Question

A heavy piece of hanging sculpture is suspended by a90-cm-long, 5.0 g steel wire. When the wind

Answer 1

As we know that fundamental frequency is given as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{m}{l}$

here we have

m = mass of wire = 5 g

l = length of wire = 90 cm

$\mu = \frac{0.005}{0.90} kg/m$

$\mu = 5.56 \times 10^{-3} kg/m$

from above formula now

$80 = \frac{1}{2(0.90)}\sqrt{\frac{T}{5.56\times 10^{-3}}}$

$144 = \sqrt{180 T}$

$T = 115.2 N$

now we know that tension is due to weight of the sculpture so we will have

$Mg = 115.2 N$

$M = 11.76 kg$

so its mass will be 11.76 kg