Adjacent antinodes of a standing wave on a string are 20.0 cm apart. A particle

Question

Adjacent antinodes of a standing wave on a string are 20.0 cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.770 cm and period 0.0765 s. The string lies along the x-axis and is fixed at x = 0.

(a) How far apart are the adjacent nodes?

                cm

(b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern?

1)    

wavelength

           cm

2) amplitude

            cm

3)

speed

            m/s

(c) Find the maximum and minimum transverse speeds of a point at an antinode.

vmax

=              m/s

vmin

=               m/s

(d) What is the shortest distance along the string between a node and an antinode?

       cm

1)    

wavelength

           cm

2) amplitude

            cm

3)

speed

            m/s

Explanation / Answer

wavelenght L = 2*20 cm= 40 cm = 0.4 m
Amplitude A =0.770 cm = 0.770x10^-2 m
peroid T =0.0765 sec
a)
the distace between adjacent nodes = 20 cm
b)
1.wavelenght L = 2*20 cm = 40 cm
2. amplitude A = 0.770 cm
speed v = w/k = (2pi/T) / (2pi/L) = L/T = 0.4/0.0765 = 5.228 m/s

c)
Vmax = WA = (2pi/T)A = (6.28/0.0765)*0.770x10^-2 = 0.632 m/s
Vmin = -0.632 m/s

d)
distance between node and antinode = L/4 = 10 cm

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