AP Physics 2 Exam Question:

A string with a length of 2 meters is fixed at both ends. The string is vibrated with a frequency of 100 Hz and creates standing waves on the string. The wave speed on the string is measured to be 200 m/s.

a) What is the wavelength of the standing wave on the string?

b) How many nodes are there in the standing wave pattern?

c) What is the distance between adjacent nodes?

Answer:

a) The wave speed (v) is given as 200 m/s and the frequency (f) is given as 100 Hz. The relationship between wave speed, frequency, and wavelength (λ) is given by the equation:

v = λf

Rearranging the equation to solve for wavelength:

λ = v / f

Substituting the given values:

λ = 200 m/s / 100 Hz

λ = 2 meters

Therefore, the wavelength of the standing wave on the string is 2 meters.

b) In a standing wave pattern, nodes are points where the amplitude of the wave is always zero. The nodes occur at intervals of half a wavelength. Hence, the number of nodes in a standing wave pattern is equal to the number of half wavelengths in the string.

Since the length of the string is given as 2 meters and the wavelength is also given as 2 meters, there is only one half wavelength in the string. This implies that there is only one node in the standing wave pattern.

Therefore, there is 1 node in the standing wave pattern.

c) The distance between adjacent nodes in a standing wave pattern is equal to one full wavelength. Given that the wavelength of the standing wave is 2 meters, the distance between adjacent nodes is also equal to 2 meters.

Therefore, the distance between adjacent nodes in the standing wave pattern is 2 meters.