Question

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that the standing wave is produced by the superposition of traveling and reflected waves, where the incident traveling waves propagate in the +x direction with an amplitude A = 2.45 mm and a speed vx = 10.5 m/s . The first antinode of the standing wave is a distance of x = 27.5 cm from the left end of the string, while a light bead is placed a distance of 13.8 cm to the right of the first antinode. What is the maximum transverse speed vy of the bead? Make sure to use consistent distance units in your calculations.


Given that the distance from the left end of the string to the first antinode is 27.5 cm , calculate the wavelength of the standing wave on the string. Remember to convert all measurements into units of meters before performing this calculation.

Answer 1

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

Answer 2

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:

$v_y=0.002m/s$

On the standing waves well-behaved, the first antinode happen one of four equal parts of a intuitiveness out completely. This means:

$\lambda=1.1 m$

This way that the connection middle from two points the wavelength and the extent of object of the strand happen:

$L=3\lambda /2$

By definition, this standing wave exist at the after second harmonious, n = 3.

Furthermore, the standing wave equating happen in this manner:

$y(x,t)=(Asin(kx))sin(wt)=Asin(w/vx)sin(wt)=((2.45*10^{-3})sin(5.7X)sin(59.94t)$

The droplet exist established on x = 0.138 m. The maximum speed exist place the derivative of the speed function equals to nothing:

$v_y=wAsin(kx)cos(wt)\\a_y=-w^2Asin(kx)sin(wt)\\a_y=0$

Find the time, we have:

$59.94t=\pi\\t=0.0524s$

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:

$v_y(x=0.138,t=0.0524)=(59.94)(2.45*10^{-3})sin((5.7)(0.138))cos((59.94)(0.0524))\\=0.002 m/s$

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