Maximum force on rope and wavefunction

• Soren4
In summary, the maximum acceleration of a rope element in a sinusoidal wave with linear density ##\mu=0.012 kg/m## is ##a_{max}=\omega^2 A=375 m/s^2##. The maximum transverse force on a piece of rope ##1 cm## long is ##F_{max, y}=m a_{max}=\mu \cdot 0.01 \cdot \omega^2 A = 0.045 N##, and the relationship between tension and force is given by ##F_y = -T \frac{\partial \xi}{\partial x}##, where ##T## is the tension of the rope and ##\frac{\partial \xi}{\partial x
Soren4

Homework Statement

A sinusoidal wave on a rope with linear density ##\mu=0.012 kg/m## is described in SI units by $$\xi(x,t)=A sin (kx-\omega t)= 0.15sin(0.8x-50t)$$
a) find the maximum acceleration of a rope element
b) find the maximum transverse force on a piece of rope ##1 cm## long
c) Show how the force depends on the tension of the rope

Wave equation

The Attempt at a Solution

a) This is simply ##a_{max}=\omega^2 A=375 m/s^2##
b) This should be ##F_{max, y}=m a_{max}=\mu \cdot 0.01 \cdot \omega^2 A = 0.045 N##
c) Tension should be related to such force by ##F_{y}=-T \frac{\partial \xi}{\partial x}## but here is the problem, since
$$\frac{\partial \xi}{\partial x}=-0.12 sin(0.8x-50t)$$

So the maximum value is ##0.12 ## and the force should be

$$F_{max,y}= 0.12 T= 0.12 \cdot \mu \cdot v^2$$

Now ##v=\frac{\omega}{k}=\frac{50}{0.8}=62.5 m/s## so it should be

$$F_{max,y}=0.09 N$$

Different from the value found in b), which is strange, did I get something wrong?

Last edited:
I think I have a clue but I'm not sure... when you say that Fy= -T de/dx ...

Isn't it Fy= -2T de/dx at the maximum point ? Since the tension acts on both sides of the rope element

And the derivative of the sine is the cosine

Last edited by a moderator:
In the equation F=ma, how exactly is F defined? Constrast that with tension in a rope.

1. What is the maximum force that a rope can handle?

The maximum force that a rope can handle depends on several factors such as the material, diameter, and condition of the rope. It is important to consult the manufacturer's guidelines for the specific rope being used.

2. How is the maximum force on a rope calculated?

The maximum force on a rope is typically calculated using the rope's breaking strength, which is the maximum amount of force that can be applied before the rope breaks. This value is usually given in pounds or kilograms and can be found in the manufacturer's specifications.

3. What is the relationship between wavefunctions and maximum force on a rope?

Wavefunctions, which are mathematical descriptions of the behavior of waves, can be used to calculate the maximum force on a rope when it is subjected to a wave-like motion. This can be useful in understanding the impact of ocean waves on a rope, for example.

4. Is there a limit to the maximum force that can be applied to a rope?

Yes, there is a limit to the maximum force that can be applied to a rope. This limit is determined by the rope's breaking strength and exceeding it can result in the rope breaking or failing in some way. It is important to always stay within the recommended maximum force for a given rope.

5. How can the maximum force on a rope be increased?

The maximum force on a rope can be increased by using a thicker or stronger rope, or by reinforcing the rope with additional materials such as metal or synthetic fibers. However, it is important to note that increasing the maximum force also increases the risk of the rope breaking and should be done with caution and proper testing.

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