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iogann1982 [59]
3 years ago
6

A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm

onic. Answer in units of Hz.
Physics
1 answer:
Galina-37 [17]3 years ago
7 0

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

v =\sqrt{\dfrac{T}{\mu}}

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}

   v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

          λ=2 L

          λ=2 x 0.577 = 1.154 m

we now,

       v = f λ

      f = \dfrac{v}{\lambda}

      f = \dfrac{1074.49}{1.154}

             f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

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