Question

A heavy piece of hanging sculpture is suspended by a90-cm-long, 5.0 g steel wire. When the wind
blows hard, the wire hums at its fundamental frequency of 80Hz. What is the mass of the
sculpture?

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

Answer 2

The mass of the heavy piece of the sculpture, which is suspended by a steel wire is 11.76 kg.

What is the fundamental frequency?

The most low-level resonating frequency, produced by a vibrating object or of a sound wave, is called the fundamental frequency.

The fundamental frequency can be given as,

$f=\dfrac{1}{2L}\sqrt{\dfrac{T}{mL}}$

Here, (L) is the length of the string, (T) is the tension in the string, and (m) is the mass of the string.

As, the mass of the steel wire is 5.0 g, and the length of the wire is 90 cm. Thus, the tension in the string for the  fundamental frequency of 80Hz can be found out using the above formula as,

$80=\dfrac{1}{2(0.90)}\sqrt{\dfrac{T}{0.005(0.9)}}\\T=115.2\rm N$

Now, the tension is the product of mass times gravitation force for the sculpture. Thus, the mass of the sculpture is,

$T=Mg\\115.2=M\times9.8\\M=11.76\rm kg$

Thus, the mass of the heavy piece of the sculpture, which is suspended by a steel wire, is 11.76 kg.

Learn more about the fundamental frequency here;

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