# Maximum force on rope and wavefunction

## 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?

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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

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haruspex