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AP Physics 2 Exam Question
A string of length L is fixed at both ends and is vibrating in its fundamental mode. The tension in the string is T and its linear density is μ. The string produces a standing wave with two antinodes, as shown in the figure below.
(a) Determine the frequency of the fundamental mode.
(b) Calculate the wavelength of the standing wave.
(c) Derive an expression for the speed of the wave on the string in terms of T and μ.
(d) Calculate the distance between the antinodes shown in the figure.
Solution
(a) To determine the frequency of the fundamental mode, we need to find the lowest possible frequency at which a standing wave with two antinodes can be produced. In the fundamental mode, the standing wave has one-half of a wavelength between the two fixed ends. Thus, the length of the string must be equal to one-half of the wavelength.
Given that the string length is L, we can write:
L = (1/2) * λ
where λ is the wavelength of the standing wave.
Rearranging the equation, we can solve for the wavelength:
λ = 2L
The frequency of a wave can be calculated using the wave equation:
v = fλ
where v is the velocity of the wave and f is the frequency.
Since we are dealing with the fundamental mode, the velocity will be the speed of the wave on the string, which will be derived in part (c).
Substituting the values into the equation, we get:
v = f * 2L
Since f is the frequency of the fundamental mode, it is the lowest possible frequency for a standing wave with two antinodes. Therefore, the frequency of the fundamental mode is given by:
f = v / (2L)
(b) The wavelength of the standing wave can be calculated using the equation derived in part (a):
λ = 2L
(c) To derive an expression for the speed of the wave on the string in terms of T and μ, we can use the equation for wave velocity on a string:
v = sqrt(T/μ)
where T is the tension in the string and μ is the linear density of the string.
(d) The distance between the antinodes shown in the figure is half of the wavelength, as mentioned in part (a). Therefore, it is equal to L.
The distance between the antinodes = L.
Markdown Format for the AP Physics 2 Exam Question and Solution:
AP Physics 2 Exam Question
A string of length L is fixed at both ends and is vibrating in its fundamental mode. The tension in the string is T and its linear density is μ. The string produces a standing wave with two antinodes, as shown in the figure below.
(a) Determine the frequency of the fundamental mode.
(b) Calculate the wavelength of the standing wave.
(c) Derive an expression for the speed of the wave on the string in terms of T and μ.
(d) Calculate the distance between the antinodes shown in the figure.
Solution
(a) To determine the frequency of the fundamental mode, we need to find the lowest possible frequency at which a standing wave with two antinodes can be produced. In the fundamental mode, the standing wave has one-half of a wavelength between the two fixed ends. Thus, the length of the string must be equal to one-half of the wavelength.
Given that the string length is L, we can write:
L = (1/2) * λ
where λ is the wavelength of the standing wave.
Rearranging the equation, we can solve for the wavelength:
λ = 2L
The frequency of a wave can be calculated using the wave equation:
v = fλ
where v is the velocity of the wave and f is the frequency.
Since we are dealing with the fundamental mode, the velocity will be the speed of the wave on the string, which will be derived in part (c).
Substituting the values into the equation, we get:
v = f * 2L
Since f is the frequency of the fundamental mode, it is the lowest possible frequency for a standing wave with two antinodes. Therefore, the frequency of the fundamental mode is given by:
f = v / (2L)
(b) The wavelength of the standing wave can be calculated using the equation derived in part (a):
λ = 2L
(c) To derive an expression for the speed of the wave on the string in terms of T and μ, we can use the equation for wave velocity on a string:
v = sqrt(T/μ)
where T is the tension in the string and μ is the linear density of the string.
(d) The distance between the antinodes shown in the figure is half of the wavelength, as mentioned in part (a). Therefore, it is equal to L.
The distance between the antinodes = L.