# Tag Info

0

Because the definition of rigid body requires that the distance between its points doesn't change. If we follow that model, only translations and rotations are possible. If we modify the model from a rigid body to an elastic body, mechanical vibrations are included. For a more complete description, heat transfer must be also included, resulting from external ...

0

we add the total translational kinetic energy and the total rotational kinetic energy of the constituent particles. Why is the total vibrational kinetic energy of the constituent particles left out? We don’t actually leave out vibrational kinetic energy if the centre of mass of the system is vibrating. This is a matter of scales at which you’re analysing ...

1

A net force or torque on a rigid body will not affect its internal energy. As it remains constant before and after the application of the force/torque, it is not relevant to the equations of motion.

1

How did the liquid levitate at 0:37 or at any one of those moments ? The liquid levitates because the gauge pressure of the gas pushing up on the bottom surface of the liquid is equal to the weight of the fluid on average. This is essentially the same reason a hovercraft levitates. I think it's because of vibrating the air in the vertical direction which ...

-1

when you apply vibrations and then flip the liquid then due to vibrations no droplets are formed and high surface tension of the liquid doesn't allow mixing of anything in it easily so some particular density may allow that to happen but some completely block the mixing of the air

0

Considering electric circuits, the Voltage is an analogy for force in mechanical systems. The inductor, capacitor and resistances are analogs to mass, spring constant and damping respectively. The frequency of forced vibration is simply the frequency of the AC voltage. At steady state after all the transients are died out, the energy for the system is ...

2

I agree that this description of forced oscillations is too short and too little - it is confusing more than explaining. Let us go point by point: First of all, let us note that the machanical system implied here is an oscillator, most likely a linear oscillator which has its own natural frequency (and probably also a damping coefficient) Periodic force ...

2

Any kind of work done on the water eventually turns into thermal agitation (heat/temperature). It first produces waves (pressure oscillations) which is nearly adiabatic during a few seconds. But as long as the waves calm down in one way or another, the energy is turned into heat. The energy required to heat 1.5 kg water from 20 degrees to 80 degrees (C) is ...

0

If you mean can a person supply sufficient pressure to raise the temperature of the water to its boiling point, the answer is no. Water is relatively incompressible. It takes an extraordinary amount of pressure to raise its temperature just a few degrees. The person could, of course, turn the paddle wheel instead of using the weight and achieve the same ...

1

It is possible, but probably not practical, for the following reasons. A near as I can tell, a light racket has the advantage that it can be wielded swiftly (low inertia) and swung up to high speed (for a good smash). Adding for example lead shot in a hollow handle will greatly increase the inertia of the racket, making it difficult to quickly maneuver into ...

0

A classical harmonic oscillator follows the differential equation: $$A\frac{d^2x}{dt^2} = -Bx$$ The solution is a sinusoidal function as: $x = K_1sin(\omega t + K_2)$ $K_1$ is the amplitude and $K_2$ the phase of the oscillation, while $$\omega = \sqrt{\frac{B}{A}}$$ is the frequency. Note that the amplitude is not determined, and theoretically any value ...

1

This concept of damping in metals is called loss factor. You can model loss factor by considering a complex Young's Modulus, such as $$E = E_0 (1+j\eta)$$ where $E_0$ is the Young's modulus of the material and $\eta$ is the loss factor. In metals $\eta$ tend to be very small (around 0.001 - 0.02).

0

Mechanical systems have properties such as mass and stiffness. There is another property derived from the first two which is called natural frequency. In general, these mechanical system can have various natural frequencies. But lets assume the representation of a single natural frequency that can be written as $$\omega_n = \sqrt{\frac{k}{m}}$$, with units ...

0

Both are similar concepts. Sound is the vibration of air particles (compression and expansion) the can reach your ears. But you can have vibration being propagated in liquids and solids as well. Some sounds are generated in structures, so the vibration of a structure is converted to sound in air — for instance, a loudspeaker.

0

The free vibration of a string can be modelled with the partial differential equation: $$T\frac{\partial^2 y(x,t)}{\partial x^2}-\rho \frac{\partial^2 y(x,t)}{\partial t^2} = 0$$ Where $T$ is the axial tension on the string, $\rho$ is the mass density, $y$ is your transverse displacement, and $x$ is the position along the string length. The solution of this ...

0

Well, i think i can help a little, see this image: If it helps you to construct a equation: $$\frac{F}{A} = Y\frac{\partial \varepsilon }{\partial x}$$ I think is enough to answer the question with this, being epsilon the wave function.

Top 50 recent answers are included