Question

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​

Answer 1

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

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 = $\sqrt { \frac{T}{ \mu } }$

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 = $\sqrt{ \frac{90 }{5.625 \ 10^{-3}} }$

         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