# Question about a stretched string's motion at the free end

• Clara Chung
In summary, there is confusion about why the equation ∂Ψ/∂x should be zero in (1.125) and whether we can apply the results of the wave equation at the free end. It is clarified that the derivation of the wave equation was done by considering all string elements that have other string elements on either side, and the imposition of boundary conditions does not violate this assumption. It is also explained that at the laterally unconstrained end, ∂Ψ/∂x=0, but not ∂Ψ/∂t. There is further discussion about the equation ∂Ψ/∂t=-c∂Ψ/∂x and its derivation, and the possibility of

#### Clara Chung

1. I don't know why ∂Ψ/∂x should be zero in (1.125). Shouldn't the tension be zero at the very last end?
2. I don't know why we can apply the results of the wave equation at the free end because we assumed there are tension both sides of the string segment during the derivation of the wave equation.

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I don’t think they are assuming that the end of the string is totally unconstrained. I think they are assuming that the string is still under tension axially, but it is free to move laterally. Imagine the end of the string attached to a small ring, with a stationary frictionless rod running vertically through the ring. So the ring can move up and down on the rod, but not left and right.

Clara Chung
For the second question, the derivation of the wave equation was done by considering all string elements that have other string elements on either side. The imposition of boundary conditions does not violate this assumption, because boundary conditions only apply to those string elements that are on the boundary, i.e., to those string elements that do not have other string elements on either side.

Clara Chung
Thanks for the answer. In this case, does it mean I can't use the formula ∂Ψ/∂x=-1/c ∂Ψ/∂t, to get the vertical velocity =0 of the segment of the end of the string?
mfig said:
For the second question, the derivation of the wave equation was done by considering all string elements that have other string elements on either side. The imposition of boundary conditions does not violate this assumption, because boundary conditions only apply to those string elements that are on the boundary, i.e., to those string elements that do not have other string elements on either side.

Clara Chung said:
Thanks for the answer. In this case, does it mean I can't use the formula ∂Ψ/∂x=-1/c ∂Ψ/∂t, to get the vertical velocity =0 of the segment of the end of the string?
There is no significant movement of material along the axis of the string. All the movement is assumed to be normal to the axis, and the material velocity is just ##\partial \psi/\partial t##. The wave velocity is c. At the laterally unconstrained end, ##\partial \psi/\partial x=0##, but not ##\partial \psi/\partial t##.

Chestermiller said:
There is no significant movement of material along the axis of the string. All the movement is assumed to be normal to the axis, and the material velocity is just ##\partial \psi/\partial t##. The wave velocity is c. At the laterally unconstrained end, ##\partial \psi/\partial x=0##, but not ##\partial \psi/\partial t##.
I know that the wave velocity is c. However if I substitute ∂ψ/∂x=0 into ∂Ψ/∂x=-1/c ∂Ψ/∂t to find the normal velocity of the segment of the string at the free end, ∂Ψ/∂t becomes zero. This means the segment is not moving at all at the free end which is a contradiction.

Clara Chung said:
I know that the wave velocity is c. However if I substitute ∂ψ/∂x=0 into ∂Ψ/∂x=-1/c ∂Ψ/∂t to find the normal velocity of the segment of the string at the free end, ∂Ψ/∂t becomes zero. This means the segment is not moving at all at the free end which is a contradiction.
Where did the equation ##\frac{\partial \psi}{\partial t}=-c\frac{\partial \psi}{\partial x}## come from?

Chestermiller said:
Where did the equation ##\frac{\partial \psi}{\partial t}=-c\frac{\partial \psi}{\partial x}## come from?
It comes from the wave equation if ψ = f(kx-wt). Does it mean I can't use the equation because ψ = f(kx-wt) + g(kx+wt)?

Clara Chung said:
It comes from the wave equation if ψ = f(kx-wt). Does it mean I can't use the equation because ψ = f(kx-wt) + g(kx+wt)?
Please show how it derives from that ewuation.

## 1. What causes a stretched string to vibrate at its free end?

A stretched string will vibrate when it is plucked or struck because of the tension and elasticity of the string. When it is plucked or struck, the string is displaced from its original position and then pulled back to its resting position. This back-and-forth movement creates vibrations, which travel through the string and produce sound waves.

## 2. Does the length of the string affect its motion at the free end?

Yes, the length of a string does affect its motion at the free end. The shorter the string, the higher the frequency of the vibrations and the higher the pitch of the sound produced. On the other hand, longer strings have lower frequencies and produce lower pitched sounds.

## 3. Can the tension of the string impact its behavior at the free end?

Absolutely. The tension of a string is directly related to its frequency and pitch. Higher tension in a string will result in higher frequencies and higher pitched sounds, while lower tension will produce lower frequencies and lower pitched sounds. Additionally, changes in tension can also affect the speed at which the waves travel through the string.

## 4. How does the material of the string influence its motion at the free end?

The material of a string also plays a role in its behavior at the free end. Different materials have different densities and elasticities, which can impact the speed at which vibrations travel through the string and the resulting sound produced. For example, a nylon string will produce a different sound than a steel string, even when they are plucked at the same tension and length.

## 5. Are there any factors that can affect the amplitude of the vibrations at the free end of a stretched string?

Yes, there are several factors that can affect the amplitude of the vibrations at the free end of a stretched string. These include the force used to pluck or strike the string, the angle at which the string is plucked, and any external forces acting on the string, such as wind or friction. Additionally, the material and thickness of the string can also impact the amplitude of the vibrations.