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

Answer:

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

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

$\frac{\lambda}{4} = 27.5 cm$

$\lambda = 110 cm$

now we have

$N = \frac{2L}{\lambda}$

$N = \frac{2(1.65)}{1.10}$

$N = 3$

now we have

$R = 2A sin(kx)$

$R = 2(3.65) sin(\frac{2\pi}{1.10}x)$

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

now at x = 13.8 cm

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

$R = 5.18 mm$

now we have

$f = \frac{v}{\lambda}$

$f = \frac{13.5}{1.1}$

$f = 12.27 Hz$

now maximum speed is given as

$v_y = R\omega$

$v_y = (5.18 \times 10^{-3})(2\pi(12.27))$

$v_y = 0.4 m/s$