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a horizontal string is stretched between two points a Distance 0.80 metre apart the tension in the string is 90 Newtonand it is mass is 4.5 g calculate the wavelength and frequency of the three lowest frequency modes of vibration of the string​

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Answer:

a) v = 126.5 m / s, b) λ₁ = 1.6 m, λ₂ = 0.8 m, λ₃ = 0.533 m, f1 = 79 Hz,

λ₂ = 0.8 m, f₃ = 237 Hz

Step-by-step explanation:

This is an exercise we are going to solve in parts, let's start by looking for the speed of the wave in the string.

v =
\sqrt { (T)/( \mu ) }

the rope tension is T = 90 N and the density can be calculated

μ = m / l

let's calculate

μ = 4.5 10⁻³ / 0.80

μ = 5.625 10⁻³ kg/m

let's calculate the speed

v =
\sqrt{ (90 )/(5.625 \ 10^(-3)) }

v = 126.5 m / s

For the second part there is a process of resonance in the string. The points where it is attached are nodes so, if L is the length of the chord

L = ½ λ 1st harmonic

L = 2/2 λ 2nd harmonic

L = 3/2 λ 3rd harmonic

L = n/2 λ n harmonic

the wavelength of the first three harmonics is requested

let's calculate

λ₁ = 2L

λ₁ = 2 0.8

λ₁ = 1.6 m

λ₂ = L

λ₂ = 0.8 m

λ₃ = ⅔ L

λ₃ = ⅔ 0.8

λ₃ = 0.533 m

To calculate the frequency we use that the speed is related to the wavelength and the frequency

v = lam f

f = v / lam

we calculate

f1 = 126.5 / 1.6

f1 = 79 Hz

f2 = 126.5 / 0.8

f2 = 158 Hz

f3 = 126.5 / 0.533

f3 = 237 Hz

User Erikreed
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