Question

A metal wire of length 20 cm and diameter 0.2 mm is stretched by a load of 2 kg weight if the density of the material of the wire is 7 upon 8 gram per centimetre cube find the fundamental frequency of vibrations of that wire

Answer 1

Answer:

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Answer 2

The fundamental frequency of vibrations of that wire is 2132 Hz

The length of the metal wire is (L) = 20 cm = 0.2 m

The diameter of the wire is (d) = 0.2 mm

∴ The radius of the wire is (r) = 0.1 mm = 10⁻⁴ m

The mass of the weight (M) = 2 kg = 2000 g

The density of the material wire (D) is = 7/8 g/cc

                                                              = 7/8 × 10⁶ g/m³

According to the formula,

The fundamental frequency,

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

Where, T is the tension of the wire, and μ is the mass density of the wire.

μ = m/L   [m = mass of the wire, V = volume of the wire]

⇒ μ = DV/L

⇒ μ = DAL/L

⇒ μ = DA

⇒ μ = πr²D

∴ n = $\frac{1}{2L}\sqrt{\frac{Mg}{\pi r^2 D}$     [T = Mg]

∴ n = $\frac{1}{2 * 0.2}\sqrt{\frac{2000*10}{3.14 * (10^{-4} )^2* \frac{7}{8}* 10^6 }$

⇒ n = 2.132 × 10³

⇒ n ≅ 2132 Hz

The fundamental frequency will be 2132 Hz.