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 = 3.65 mm and a speed vx = 13.5 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?

Answer 1

The maximum transverse speed of the bead is 0.4 m/s

Step-by-step explanation:

As we know that the Amplitude of the travelling wave is

A = 3.65 mm

Now the speed of the travelling wave is

`v_x = 13.5 m/s`

now we know that distance of first antinode from one end is 27.5 cm

so length of the loop of the standing wave is given as

`(\lambda)/(4) = 27.5 cm`

`\lambda = 110 cm`

now we have

`N = (2L)/(\lambda)`

`N = (2(1.65))/(1.10)`

`N = 3`

now we have

`R = 2A sin(kx)`

`R = 2(3.65) sin((2\pi)/(1.10)x)`

`R = 7.3 sin(1.82 \pi x)`

now at x = 13.8 cm

`R = 7.3 sin(1.82 \pi (0.138))`

`R = 5.18 mm`

now we have

`f = (v)/(\lambda)`

`f = (13.5)/(1.1)`

`f = 12.27 Hz`

now maximum speed is given as

`v_y = R\omega`

`v_y = (5.18 * 10^(-3))(2\pi(12.27))`

`v_y = 0.4 m/s`