AP Physics 1 Exam Question:

A string with a length of 2 meters is fixed at both ends. The speed of wave propagation on this string is 10 m/s. Two waves, each with an amplitude of 0.1 meters and a frequency of 2 Hz, are traveling in opposite directions on the string.

a) Calculate the wavelength of each wave.

b) What is the phase difference between the two waves?

c) Determine the resulting interference pattern produced by these waves.

Answer:

a) To find the wavelength of each wave, we can use the equation:

wavelength = speed / frequency

Given: speed of wave propagation = 10 m/s and frequency = 2 Hz

Plugging in the values, we get:

wavelength = 10 m/s / 2 Hz = 5 meters

Therefore, the wavelength of each wave is 5 meters.

b) The phase difference between two waves traveling in opposite directions is half the wavelength. In this case, the phase difference would be half of 5 meters, which is 2.5 meters.

c) The interference pattern produced by these waves can be determined by considering the superposition of two waves.

If the phase difference between the waves is a multiple of the wavelength (nλ; n is an integer), constructive interference occurs, resulting in a standing wave with maximum amplitude.

If the phase difference between the waves is an odd multiple of half the wavelength ((2n+1)(λ/2); n is an integer), destructive interference occurs, resulting in a standing wave with nodes (points of minimum amplitude).

In this case, since the phase difference is 2.5 meters, which is not a multiple of the wavelength (5 meters), we will have both constructive and destructive interference happening simultaneously.

The resultant interference pattern will consist of alternating regions of constructive interference (maximum amplitude) and destructive interference (minimum amplitude). This pattern would appear as a series of equally spaced nodes and antinodes along the string, forming a standing wave.