# Tag Info

1

Let one end of a very long string is being oscillated transversely so as to generate a sinusoidal wave traveling out along the string. In order to set up a wave on a stretched string, the driving force at the end of the string provides energy. This energy is not retained at the source; it flows along the string at the wave speed. The string transports ...

1

No. Energy conservation always applies. The elastic potential energy will be maximum at a wavetop, since here the rope is stretched the most, $U=½kx^2$. The transverse velocity and thus the kinetic energy is zero at this point $K=½mv^2$ since this part of the rope stops and starts moving back again. $$E_{before}=E_{after} \implies K_1+U_1=K_2+U_2$$ Energy ...

-3

To calculate the trajectory means to find the relation between $v$, $\omega$, $A$, and $x$ in the $v$-$x$ coordinate system which leads to the ellipsoid curve.

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The answer depends on how particular you want to be about it. Upon first examination you might say that in the Earth's conservative gravitational field, that the work must be the same, since the start and endpoints are the same and we have ignored any non-conservative forces (e.g. friction, air resistance, etc.). This is more than likely the response that ...

2

The other answers here are in the spirit of what you can do, but allow me to elaborate a little more. To understand if the trajectory of the movement under a potential $V$ is stable or not you have to understand what this stability means. The most simple example is the harmonic oscillation- $V=-{1 \over 2}kx^2$.In Newtonian mechanics, for a point of ...

0

Once you have the value $$dV_{eff}\over dr$$ This is the minus of the effective force i.e.: $$F_{eff}=-\frac{dV_{eff}}{ dr}$$ If we differentiate this again we get: $$\frac{dF_{eff}}{dr}=-\frac{d^2V_{eff}}{ dr^2}$$ If this is negative then at any small permutation around that radius will feel a force back to that radius and thus it is a stable orbit. If it ...

0

One method for stability anayisis in a potential is to consider $r$ and $r+\epsilon$ where $\epsilon$ is a small number. If you can solve the dynamics analytically or run a numerical computer simulartion for your object with both $r$ and $r+\epsilon$ and observe with time if $\epsilon$ gets smaller or larger that will give you a clue as to whether the orbit ...

1

It seems, based on the comments above, that you have figured it out. Just for closure, I am writing the steps out. If you had just a parabolic potential well, $V(x) = ax^2$, you could get the period quite easily - for a given mass $m$, the frequency would be $$\omega = \sqrt{\frac{2\alpha}{m}}\\ T = \frac{2\pi}{\omega} = \pi \sqrt{\frac{2m}{\alpha}}$$ For ...

0

Notice that what you need to do to calculate the whole potential energy is to just count the pair of charges and add the potential between them. What I mean by that is the following: On a hexagon you have 6 vertices with length say $a$, which corresponds to a potential $\frac{kq^2}{a}$ each, 6 diagonals with length $\sqrt 3 a$, which corresponds to a ...

1

I have highlighted some key word lacking in your revision. Also, work has a very specific definition. The difference in gravitational potential difference between $\vec{r}_1$ and $\vec{r}_2$ is the negative of the work done on a unit mass by the external gravitational field as the unit mass moves from $\vec{r}_1$ to $\vec{r}_2$. As an example, consider a ...

1

Words are imprecise. Your wording is much less imprecise than the one from the textbook. As the work is positive or negative (depending on the direction taken), there is no need to define a "positive direction", but the potential difference depends on the order of $\vec r_1$ and $\vec r_2$ (that is: if the potential difference moving from 1 to 2 is positive, ...

1

1/f spectra have the unique distinction of being "scale invariant" in the sense that the energy in an interval df is proportional to df. The 1/f spectra in fact have the property that the in an interval with width df available energy is proportional to df but not with f. There, namely "scale invariant" attribute for. It is not the energy, but the signal ...

0

Potential has a negative sign as work has to be done against the external force to bring the mass from infinity to the point (definition of potential )

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Because of a convention wherein zero gravitational potential is said to be at infinity. See Wikipedia: $V(x) = \frac{W}{m} = \frac{1}{m} \int\limits_{\infty}^{x} F \ dx = \frac{1}{m} \int\limits_{\infty}^{x} \frac{G m M}{x^2} dx = -\frac{G M}{x}$ "By convention, it is always negative where it is defined, and as x tends to infinity, it approaches zero." ...

1

Hmm. I think there should be a minus sign and it doesn't matter where you set the zero of potential. Gravitational field strength is the negative gradient of the potential. For a spherically symmetric field $$g(r) = -\frac{dV}{dr}.$$ If $$V = -\frac{GM}{r} + V_0,$$ where $V_0$ is an arbitrary constant, then $$g = -\frac{GM}{r^2}.$$ i.e. your book is ...

3

The difference probably is that the graph for the gravitational potential is the one for a spherical mass distribution (or a sphere with a certain mass if you wish) and the electric one is given for a point charge. You could also draw the gravitational potential for a point mass, then it would look equivalent to your electrical potential, or the other way ...

2

So the net work on the object is zero and it doesn't gain any energy. The object obviously did gain energy. The object's potential energy increased by $mgh$, and its kinetic energy didn't change. So what's going on? Is the work-energy principle wrong? The answer is no, the work-energy principle is not wrong. The work energy principle merely says that ...

0

Gravitational weight is a conservative force so work done against it is entirely equal to the change in potential between the start and end points (in the absence of dissipative forces like friction). The internal energy of the mass does not change as it is moved upwards and your assessment of the energy considerations is correct. Consider alternatively the ...

-1

Gravity doesn't do -mgh work on the object while it is lifted, gravity converts the potential energy gained while lifting the object (mgh) into kinetic energy (1/2 mv^2) after it's dropped.

2

The total work done on the object is the change in kinetic energy: $W_{total} = \Delta E_K$* While the gravitational potential energy of the object is: $U_G = mgh$ So, although it costs energy to lift the object up, the total work done on it is $0$ because both at the beginning and at the end it has no kinetic energy (\$v_{i,f}=0 \rightarrow E_{K_{i,f}}=0 ...

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