Respuesta :
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,
[tex]v =\sqrt{\dfrac{T}{\mu}}[/tex]
μ is the mass per unit length
μ = 0.325 x 10⁻³/ 0.577
μ = 0.563 x 10⁻³ Kg/m
now,
[tex]v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}[/tex]
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
[tex]f = \dfrac{v}{\lambda}[/tex]
[tex]f = \dfrac{1074.49}{1.154}[/tex]
f = 931.1 Hz
The frequency for first harmonic is equal to f = 931.1 Hz
The frequency of first harmonic will be "931.1 Hz". To understand the calculation, check below.
Tension and Frequency
According to the question,
Wire's mass, m = 0.325 g
Stretch's length, L = 57.7 cm or,
= 0.577 m
Wire's tension, T = 650 N
Mass per unit, μ = [tex]\frac{0.325\times 10^{-3}}{0.577}[/tex]
= 0.563 × 10⁻³ Kg/m
We know the relation,
→ v = [tex]\frac{T}{\mu}[/tex]
By substituting the values, we get
= [tex]\frac{650}{0.563\times 10^{-3}}[/tex]
= 1074.49 m/s
Now,
→λ = 2 L
= 2 × 0.577
= 1.154 m
hence,
The frequency will be:
→ f = [tex]\frac{v}{\lambda}[/tex]
By substituting the values,
= [tex]\frac{1074.49}{1.154}[/tex]
= 931.1 Hz
Thus the above response is correct.
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