Respuesta :
Answer:
On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means
[tex]\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m[/tex]
This means that the relation between the wavelength and the length of the string is
[tex]3\lambda/2 = L[/tex]
By definition, this standing wave is at the third harmonic, n = 3.
Furthermore, the standing wave equation is as follows:
[tex]y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)[/tex]
The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.
[tex]v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)[/tex]
[tex]a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0[/tex]
For this equation to be equal to zero, sin(59.94t) = 0. So,
[tex]59.94t = \pi\\t = \pi/59.94 = 0.0524~s[/tex]
This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:
[tex]v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s[/tex]
In this exercise we have to use the knowledge of mechanics to be able to calculate the wavelength, so we can say that it will correspond to:
[tex]v_y=0.002m/s[/tex]
On the standing waves well-behaved, the first antinode happen one of four equal parts of a intuitiveness out completely. This means:
[tex]\lambda=1.1 m[/tex]
This way that the connection middle from two points the wavelength and the extent of object of the strand happen:
[tex]L=3\lambda /2[/tex]
By definition, this standing wave exist at the after second harmonious, n = 3.
Furthermore, the standing wave equating happen in this manner:
[tex]y(x,t)=(Asin(kx))sin(wt)=Asin(w/vx)sin(wt)=((2.45*10^{-3})sin(5.7X)sin(59.94t)[/tex]
The droplet exist established on x = 0.138 m. The maximum speed exist place the derivative of the speed function equals to nothing:
[tex]v_y=wAsin(kx)cos(wt)\\a_y=-w^2Asin(kx)sin(wt)\\a_y=0[/tex]
Find the time, we have:
[tex]59.94t=\pi\\t=0.0524s[/tex]
This exist moment of truth when the speed is maximum. So, the maximum speed maybe raise by stop up existing time into the velocity function:
[tex]v_y(x=0.138,t=0.0524)=(59.94)(2.45*10^{-3})sin((5.7)(0.138))cos((59.94)(0.0524))\\=0.002 m/s[/tex]
See more about wavelength at brainly.com/question/7143261
