# Tag Info

## New answers tagged string

5

(a) Write down the potential energy of the rope as the function $y(x)$. You're almost right, up to a minus sign in the limits of the integral: $$V=\dfrac{mg}{l}\int_{-x_0}^{d-x_0}y(x)\sqrt{1+y'^2}\ dx$$ (b) Since the problem is static, interpret the potential energy as the Lagrangian and find the Lagrangian density. The Lagrangian is usually given ...

1

The best way to define tension is as the one dimensional version of the stress tensor. So, you can define the tension at some point P in the string as the force at which the part of the string on one side of P pulls on the other side of P. The direction of this force then depends on which sides you are considering, so you should make some choice here and ...

0

I don't think you are doing anything wrong. You're simply forgetting that your reflexion co-efficient, calculated as a phase, assumes sinusoidal excitation and thus, tacitly, you've assumed that the wave system is of infinite extent in time. In particular, the system has reached a steady state. What energy is going into the mass is periodically coming out ...

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For linear deformations, when a material is compressed/stretched in one direction, it usually tends to expand/contract in the other two directions perpendicular to the direction of compression/stretching. This is called the Poisson effect, and the Poisson ratio is a measure of this effect. For some materials the Poisson ratio can be close to 0.5 (rubber) ...

2

The article you refer to is talking about the speed of sound (or speed of longitudinal wave vibrations) within a material and how that relates to volumetric mass density. If you were transmitting sound from one end of a guitar string to another, this would be relevant. But the sound produced from a string is not related to the speed that it travels within ...

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The characteristic speed, $S$ of transverse waves on a taut string or wire with tension $T$ and density per unit length, $\rho$ is $$S=\sqrt \frac T\rho$$ So indeed as you remove wire from play by wrapping it around the capstan, you decrease the wires density and increase tension. This increases velocity. and since $$\omega=\frac v\lambda$$ frequency ...

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Your intuition is right: the density of the string goes down a little bit when you increase the tension. HOWEVER: the wave in a string is a transverse wave which depends on the tension and the mass per unit length. If you double the tension the mass per unit length goes down by a small amount (the string gets a bit "thinner" because it gets longer) . Both ...

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Look at the equation. There are two components - one describes variation of the max amplitude with position (the $\sin (kx)$ part) and the other describes the variation with time ($\cos(\omega t)$). The antinodes must occur where the max amplitude is as big as possible (so where $|\sin(kx)|=1$). This will happen at multiple locations (there will be a ...

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Taking the derivative of equation $y$ (as the hint states) and solving $x$ such that it is zero will give you your answer: $$\frac{dy}{dx}=\frac{d}{dx}\left(2A\sin\left(kx\right)\cos\left(\omega t\right)\right)=0$$

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