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Oscillation of a 260 Hz tuning fork sets up standing waves in a string clamped at both ends. The wave speed for the string is 640 m/s. The standing wave has four loops and an amplitude of 3.1 mm.
(a) What is the length of the string?
(b) Write an equation for the displacement of the string as a function of position and time. Round numeric coefficients to three significant digits.

Respuesta :

Answer

given,

frequency of the tuning fork = 260 Hz

speed of wave in the string = 640 m/s

number of loop = n = 4

Amplitude = 3.1 mm

a) wavelength of the spring

 [tex]\lambda = \dfrac{v}{f}[/tex]

 [tex]\lambda = \dfrac{640}{260}[/tex]

 [tex]\lambda =2.46\ m[/tex]

we know length of string

 [tex]L = \dfrac{n\lambda}{2}[/tex]

 [tex]L = \dfrac{4\times 2.46}{2}[/tex]

 [tex]L =4.92\ m[/tex]

b) angular frequency of standing waves

  ω = 2 π f  = 2 π x 260

  ω = 520 π rad/s

  wave number

  [tex]k =\dfrac{2\pi f}{v}[/tex]

  [tex]k =\dfrac{2\pi\times 260}{600}[/tex]

       k = 2.723 rad/m

 y (x,t) = Ym sin(kx)cos(ωt)

 y (x,t) = 3.1 sin (2.723 x) cos(520 π t)

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