Question

Adjacent antinodes of a standing wave on a string are 15.0 cmapart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 cm and period 0.0750 s. The string lies along the +x-axis and is fixed at x=0.

Part D
What is the speed of the two traveling waves that form this pattern?

v = 400 m/s

Answer 1

The speed of each of the two traveling waves that form the standing wave pattern is 4.00 m/s, calculated using the distance between adjacent antinodes and the period of the oscillation.

Step-by-step explanation:

The speed of the two traveling waves that form a standing wave pattern can be determined by using the given distance between adjacent antinodes, which corresponds to half of the wavelength (λ/2), and the period (T) of the particles' oscillation at the antinodes.

First, we find the full wavelength (λ) by doubling the distance between antinodes,

so λ = 15.0 cm * 2

= 30.0 cm

= 0.30 m.

Next, using the period (T = 0.0750 s), we calculate the frequency (f) of the oscillation with the formula f = 1/T.

This gives us f = 1/0.0750 s

≈ 13.33 Hz.

The wave speed (v) is the product of the frequency and the wavelength, v = fλ.

So, v = 13.33 Hz * 0.30 m

= 4.00 m/s.

This is the speed of each of the two waves that form the standing wave pattern.