Question:
5. A rope that has uniform thickness is hanging vertically from an oscillator. When the oscillator is switched on and adjusted to a certain frequency, the rope forms a standing wave.
(a) Let P and Q be two points on the rope as shown above. Explain why the tension at P is greater than the tension at Q.
(b) Explain whether increasing the tension in a rope can increase the speed of waves (or wave speed) along the rope. You may use diagrams or/and equations to explain the standing wave as shown above.
Scoring Guidelines:
(a) Indicate that there is more weight of a rope below one point than the other. [1 mark] Indicate that the tension at a point supports the weight of the rope below the same point. [1 mark]
(b) Indicate that the wavelength is either longer near the top of the rope or shorter near the bottom. [1 mark]
Indicate that the frequency of the wave is the same throughout the rope. [1 mark]
Use an equation to conclude that wave speed is greater near the top of the rope or lesser near the bottom, based on a difference in wavelength. [1 mark]
Indicate that the tension is greater near the top of the rope or lesser near the bottom. [1 mark]
Provide a well-structured answer [1 mark].
(https://secure-media.collegeboard.org/digitalServices/pdf/ap/apcentral/ap16_physics_1_q5.pdf)
Possible answers:
(a) The small section of rope at the location P supports the length of rope immediately below P. The rope at P supports more weight as compared to the rope at Q and thus, the tension at P is higher.
(b) The wavelength near the top of the rope is observed to be longer as compared to the wavelength near the bottom.
By using the equation v = fλ, the wave speed (v) is greater at a higher location because the frequency (f) is the same throughout the rope and the wavelength (λ) is longer near the top. Thus, we can deduce that higher tension near the top of the rope results in higher wave speed.
Alternatively, we can explain the wave speed by using the equation, v = √(FT/μ), in which FT is the tension of the string and μ is the linear mass density (or mass per unit length) of the string.
Feynman’s insights or goofs?:
Feynman provides some insights on the wave equation, ∂2χ/∂x2 = (1/cs2) ∂2χ/∂t2, which describes the behavior of sound in matter. In his own words, “If we take a thin slab of air of length Δx and of unit area perpendicular to x, then the mass of air in this slab is ρ0Δx and it has the acceleration ∂2χ/∂t2, so the mass times the acceleration for this slab of matter is ρ0Δx(∂2χ/∂t2). (It makes no difference for small Δx whether the acceleration ∂2χ/∂t2 is evaluated at an edge of the slab or at some intermediate position.) If now we find the force on this matter for a unit area perpendicular to x, it will then be equal to ρ0Δx(∂2χ/∂t2). We have the force in the +x-direction, at x, of amount P(x, t) per unit area, and we have the force in the opposite direction, at x+Δx, of amount P(x+Δx, t) per unit area: P(x, t) − P(x+Δx, t) = −(∂P/∂x)Δx... (Feynman et al., 1963, section 47–3 The wave equation).” That is, the wave equation can be understood and derived by using Newton’s second law of motion.
In short, Feynman mentions that “sound is a branch of mechanics, and so it is to be understood in terms of Newton’s laws (Feynman et al., 1963, section 47–1 Waves).” Moreover, he suggests that the physics of sound waves in a gas involves three steps: First, the movement of the gas molecules changes the density of the gas. Second, the change in density of the gas corresponds to a change in the gas’s pressure. Third, the differences in pressure of the gas result in movement of the gas molecules. In other words, the movement of the gas molecules is closely related to the differences in pressure of the gas.
Similarly, we can derive the wave equation for a vibrating string in 3 steps as shown below.
Step 1: By using Newton’s second law of motion (F = ma) on a small element of the string moving in the vertical direction, FT = m d2y/dt2 = μΔx d2y/dt2 in which μ is mass per unit length of the string.
Step 2: The tension of the string acting on the small element of the string moving in the vertical direction, FT sin θ2 – FT sin θ1 ≈ FT tan θ2 – FT tan θ1
(Note: sin θ ≈ tan θ for a small angle, θ.)
In other words, the net vertical force acting on the small element of the string = FT(Δy2/Δx – Δy1/Δx) = FT (slope at x + Δx) – FT (slope at x)
Step 3: By equating the two expression: FT(Δy2/Δx – Δy1/Δx) = μΔx d2y/dt2
Thus, FT(Δy2/Δx – Δy1/Δx)/Δx = μ d2y/dt2
As Δx approaches zero, we have FT∂2y/∂x2= μ ∂2y/∂t2.
The wave equation of the string can be expressed as ∂2y/∂x2 = (1/[FT/μ]) ∂2y/∂t2 and the wave speed of the string can be represented by the equation, v = √(FT/μ).
In summary, if we represent the wave equation as FT∂2y/∂x2= μ ∂2y/∂t2, the right-hand side of the equation μ ∂2y/∂t2 is about the vertical acceleration of an element of a string times its mass per unit length, whereas the left-hand side of the equation (FT∂2y/∂x2) is related to the rate of change of the slope of the string with respect to distance times the tension of the string.
Importantly, we can explain that an increase in the tension (FT) of a rope can increase the wave speed of the rope by using the wave equation or v = √(FT/μ). Alternatively, we can apply Newton’s second law of motion on a small element of the vertical rope, FT = m d2x/dt2 = μΔy d2x/dt2 in which μ is the mass per unit length of the rope. In other words, the increase in the tension of the rope can increase the speed of waves along the rope.
Reference:
1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
5. A rope that has uniform thickness is hanging vertically from an oscillator. When the oscillator is switched on and adjusted to a certain frequency, the rope forms a standing wave.
(a) Let P and Q be two points on the rope as shown above. Explain why the tension at P is greater than the tension at Q.
(b) Explain whether increasing the tension in a rope can increase the speed of waves (or wave speed) along the rope. You may use diagrams or/and equations to explain the standing wave as shown above.
Scoring Guidelines:
(a) Indicate that there is more weight of a rope below one point than the other. [1 mark] Indicate that the tension at a point supports the weight of the rope below the same point. [1 mark]
(b) Indicate that the wavelength is either longer near the top of the rope or shorter near the bottom. [1 mark]
Indicate that the frequency of the wave is the same throughout the rope. [1 mark]
Use an equation to conclude that wave speed is greater near the top of the rope or lesser near the bottom, based on a difference in wavelength. [1 mark]
Indicate that the tension is greater near the top of the rope or lesser near the bottom. [1 mark]
Provide a well-structured answer [1 mark].
(https://secure-media.collegeboard.org/digitalServices/pdf/ap/apcentral/ap16_physics_1_q5.pdf)
Possible answers:
(a) The small section of rope at the location P supports the length of rope immediately below P. The rope at P supports more weight as compared to the rope at Q and thus, the tension at P is higher.
(b) The wavelength near the top of the rope is observed to be longer as compared to the wavelength near the bottom.
By using the equation v = fλ, the wave speed (v) is greater at a higher location because the frequency (f) is the same throughout the rope and the wavelength (λ) is longer near the top. Thus, we can deduce that higher tension near the top of the rope results in higher wave speed.
Alternatively, we can explain the wave speed by using the equation, v = √(FT/μ), in which FT is the tension of the string and μ is the linear mass density (or mass per unit length) of the string.
Feynman’s insights or goofs?:
Feynman provides some insights on the wave equation, ∂2χ/∂x2 = (1/cs2) ∂2χ/∂t2, which describes the behavior of sound in matter. In his own words, “If we take a thin slab of air of length Δx and of unit area perpendicular to x, then the mass of air in this slab is ρ0Δx and it has the acceleration ∂2χ/∂t2, so the mass times the acceleration for this slab of matter is ρ0Δx(∂2χ/∂t2). (It makes no difference for small Δx whether the acceleration ∂2χ/∂t2 is evaluated at an edge of the slab or at some intermediate position.) If now we find the force on this matter for a unit area perpendicular to x, it will then be equal to ρ0Δx(∂2χ/∂t2). We have the force in the +x-direction, at x, of amount P(x, t) per unit area, and we have the force in the opposite direction, at x+Δx, of amount P(x+Δx, t) per unit area: P(x, t) − P(x+Δx, t) = −(∂P/∂x)Δx... (Feynman et al., 1963, section 47–3 The wave equation).” That is, the wave equation can be understood and derived by using Newton’s second law of motion.
In short, Feynman mentions that “sound is a branch of mechanics, and so it is to be understood in terms of Newton’s laws (Feynman et al., 1963, section 47–1 Waves).” Moreover, he suggests that the physics of sound waves in a gas involves three steps: First, the movement of the gas molecules changes the density of the gas. Second, the change in density of the gas corresponds to a change in the gas’s pressure. Third, the differences in pressure of the gas result in movement of the gas molecules. In other words, the movement of the gas molecules is closely related to the differences in pressure of the gas.
Similarly, we can derive the wave equation for a vibrating string in 3 steps as shown below.
Step 1: By using Newton’s second law of motion (F = ma) on a small element of the string moving in the vertical direction, FT = m d2y/dt2 = μΔx d2y/dt2 in which μ is mass per unit length of the string.
Step 2: The tension of the string acting on the small element of the string moving in the vertical direction, FT sin θ2 – FT sin θ1 ≈ FT tan θ2 – FT tan θ1
(Note: sin θ ≈ tan θ for a small angle, θ.)
In other words, the net vertical force acting on the small element of the string = FT(Δy2/Δx – Δy1/Δx) = FT (slope at x + Δx) – FT (slope at x)
Step 3: By equating the two expression: FT(Δy2/Δx – Δy1/Δx) = μΔx d2y/dt2
Thus, FT(Δy2/Δx – Δy1/Δx)/Δx = μ d2y/dt2
As Δx approaches zero, we have FT∂2y/∂x2= μ ∂2y/∂t2.
The wave equation of the string can be expressed as ∂2y/∂x2 = (1/[FT/μ]) ∂2y/∂t2 and the wave speed of the string can be represented by the equation, v = √(FT/μ).
In summary, if we represent the wave equation as FT∂2y/∂x2= μ ∂2y/∂t2, the right-hand side of the equation μ ∂2y/∂t2 is about the vertical acceleration of an element of a string times its mass per unit length, whereas the left-hand side of the equation (FT∂2y/∂x2) is related to the rate of change of the slope of the string with respect to distance times the tension of the string.
Importantly, we can explain that an increase in the tension (FT) of a rope can increase the wave speed of the rope by using the wave equation or v = √(FT/μ). Alternatively, we can apply Newton’s second law of motion on a small element of the vertical rope, FT = m d2x/dt2 = μΔy d2x/dt2 in which μ is the mass per unit length of the rope. In other words, the increase in the tension of the rope can increase the speed of waves along the rope.
Reference:
1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
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