Angular frequency ω of the mass's oscillations?
uncomplicated Harmonics concern y = A*sin(w*t) y = 0.232 m * sin (2.80 one rad/s * t) w = sqrt(ok/m) ==> m = ok/w^2 = 35.2 N/m / (2.80 one rad/s)^2 = 4.40 six kg The trick to this concern is they set t = 0 sec y = .232 m The SH equation instruments the displacement = 0 at t = 0 sec, so we ought to locate the offset y = 0.232 m * sin (2.80 one rad/s * t + C) clean up for C .232 m = 0.232 m * sin(2.80 one rad/s * 0 s + C) a million = sin(0 + C) C = asin(a million) = pi/2 rad or ninety deg So, our equation for this concern is y = .232 m * sin(2.80 one rad/s * t + pi/2) For a spring concern, the optimal spring capacity happens on the precise of the course, so Spring = a million/2*ok*x^2 Spring = a million/2 * 35.2 N/m * (.232 m) ^2 = 0.947 J in the process the parth, the Spring capacity gets transferred into Kinetic capacity, with the optimal occuring at y = 0 m. Any combination there in is comparable to the Spring capacity Max. The trick to this concern is to locate the rate on the element t = a million.40 two s and then looking the Kinetic capacity and subtracting that from the Max Spring capacity. i'm undecided while you're employing calculus on your type, yet this is a thank you to clean up for the rate employing it. v = dy/dt = 2.80 one a million/s * .232 m * cos ( 2.80 one rad/s * t + pi/2 ) v = 0.652 m/s * cos ( 2.80 one rad/s * t + pi/2 ) Plug in t = a million.40 two s v = 0.652 m/s * cos (2.80 one rad/s * a million.40 two s + pi/2) = 0.489 m/s Kinetic capacity = a million/2*m*v^2 = a million/2* 4.40 six kg * (0.489 m/s)^2 = 0.534 J Spring capacity = Max Spring capacity - Kinetic capacity = 0.947 J - 0.534 J = 0.413 J
Find the angular frequency of oscillation when the period is 0.069 s.?
A) T = 1/frequency B) w = 2*pi/T
Need help with angular frequency of oscillation of an object?
Two identical thin rods, each of mass m and length L, are joined at right angles to form an L-shaped object. This object is balanced on top of a sharp edge (Figure 1) . If the object is displaced slightly, it oscillates. Assume that the magnitude of the acceleration due to gravity is g. Find ω, the angular frequency of oscillation of the object. Your answer for the angular frequency may contain the given variables m and L as well as g.
Find the angular frequency of oscillation? please help thanks!?
When balanced the two legs make angle "a" = 45 deg. with the vertical. When set into osscilation the legs will move thru small angle "e" about the equilibrium angle. The equations of motion are; Net Torque on legs = - Moment of Inetia of legs x angular acceleration When system is pulled aside to angle (a + e) on one leg and angle (a-e) on the other the net torque due to gravity is; T(net) = (M/2)g(L/2)Sin(a + e) - (M/2)g(L/2)Sin(a - e) =(M/2)g(L/2)2Cos(a)Sin(e) Where I used double angle identity on the Sines. Also, M = total mass Now the moment of inertia about one end of a rod is (1/3)ML^2 . So for the two rods; I = (1/3)(M/2)L^2 + (1/3)(M/2)L^2 = (M/3)L^2 The equation of motion is; (1/2)MgLCos(a)Sin(e) = - (M/3)L^2(aa) For small angles ; Sin(e) ~ e This is a differential equation ,usually written as; (aa) + [(3/2)(g/L)Cos(a)]e = 0 Now (aa) is the second time derivative of "e" , So this is a classic harmonic oscillator equation with frequency; angular frequency = SqRt[(3/2)(g/L)Cos(a)] in radians/sec with Cos(a) = (1/2)SqRt(2)
Frequency of oscillation of mass on spring?
The frequency of an object in simple harmonic motion is sqrt(k/m) / (2 * pi) We know that the force exerted on the spring is represented by the formula F = kx where k = spring constant x = elongation (distance from equilibrium position) Since the only force acting on the spring is the weight of the mass we can say that mg = kx We want to solve for k since we have all the other information so k = mg / x = 2 * 9.8 / (0.5) = 39.2 N/m *We use 0.5 for elongation because k is conventionally represented as N/m so we need to convert cm to m* Now we have all the information we need to plug into our frequency formula f = sqrt(k/m) / (2 * pi) = sqrt(39.2 / 2) / (2 * pi) = 0.704609 Hz
What is the frequency of oscillation of simple pendulum mounted in a cable that is falling freely under gravity?
The answer is zero - the pendulum will not be in oscillation.Even if the pendulum had been in oscillation before the commencement of the cabin’s free fall, this oscillation will cease very quickly once the free fall begins.Hitherto, the pendulum swing was due to three identifiable forces: 1) The downwardly directed force of gravity 2) The tension in the string 3) The angular momentum of the pendulum.At the commencement of the cabin’s free fall, whereas the downwardly directed force of gravity still acts on the pendulum, the tension force in the string disappears because the cabin roof, presumably from which the pendulum hangs, is in free fall also and thus a “reaction” pull by the string becomes inexistent since there is an inexistent pull by the pendulum mass also, as both objects have zero motion relative to each other, nor is there a dynamic equilibrium of forces acting between them.The angular momentum earlier, was a resultant of the gravitational pull on the pendulum against the ever present tension on the string (drivable by the parallelogram of forces theorem and ever changing its direction as a new component, with every new component remaining perpendicular to the tension force in the string at any point in time along the swing).With the downward gravitational pull and the tension in string effectively neutralized by the free fall of the cabin, angular momentum also ceases and with it any further swinging motion of the pendulum.
What will happen if sun start oscillating up and down with some large angular frequency omega?
Just a gentle reminder that the Earth is orbiting the sun, not the other way around. If for some reason the sun appeared to be moving up and down in our sky, it seems more likely that the tilt of the Earth is moving up and down rapidly (though it is hard to explain either situation really, it just seems less likely that the much more massive object would be moving in such a way).If this were to happen, I do not see why we would see any particularly interesting gravity wave phenomena as mentioned in the other answers; we would probably feel such a movement happening though, and it would probably get very messy very quickly all over the world.If the sun instead were to start oscillating in space noticeably with respect to the solar plane, that would probably indeed create some interesting phenomena regarding gravity. It probably also would not be good for those living on the Earth.
What is the relation between frequency and current?
It all depends on the basis on which you are talking about frequency and current. Since you have not defined the involvement of both parameters frequency and current in some process, the processes are categorised below.See below, in which you are interested.Electrical circuitsIn AC electrical circuits, there are many components where many factors like current, voltage, impedance and frequency are correlated. Talking about inductive load, it's inductive reactance is directly proportional to frequency of operation and thus since current is inversely proportional to the inductive reactance. So as the frequency increases, the inductive reactance also increases which in turns decreases the current through the load.If we consider capacitor, then in this case as the frequency increases, the capacitive reactance decreases which thus increases the current through capacitor. In case of resistive load, the current is only dependent on its resistance, it doesn't depend on the frequency of AC supply.Photoelectric effectIt's a phenomenon by which illumination of radiation(light) results in ejection of electrons from the surface of a material which thus gives rise to current. In this case, the current is independent of the frequency of illuminated light but it depends on the intensity of light used in the process. The frequency of light controls the velocity of electrons towards the collector plate and thus the stopping potential. As the frequency increases, the stopping potential also increases. While more intensity of light results in more number of photons hitting the surface and this enhances the flow of large number of electrons towards the collector plate. In this way the current is dependent on the intensity of illuminated light.Thanks for the question.
Physics Oscillation question?
A block on a spring is pulled to the right and released at t = 0 s. The block passes 3.27 cm at t = 0.592 s and it passes -3.27 cm at 0.875 s. What is the angular frequency of the oscillation (in rad/s)? Hint: cos(π - θ) = -cos(θ). Also, what is the amplitude(in cm)? This is the only question I have been unable to answer. I'm just unsure how to use the hint to help answer it. I considered obtaining average velocity from the data but I'm not sure where to go from there. It doesn't seem like too difficult a question though. Someone else may have an easier time. I appreciate your assistance.
Why is the reduced Planck constant useful in angular frequency?
If you analyse the units, you will find that Planck’s constant [math]h[/math] has the units of angular momentum.You have [math][E] = [M][D]^2[T]^{-2}[/math] and [math][f]=[T]^{-1}.[/math]So it follows that [math][h]=[M][D]^{2}[T]^{-1}=[L=mvr].[/math]So you might well guess that angular momentum will be quantized in terms of units of [math]h[/math], and indeed this turns out to be the case. But since the quantization of angular momentum involves the quantization of rotational motion, it turns out that it is necessary to divide Planck’s constant by [math]2\pi[/math] to get the quantization condition, which in Bohr’s case was [math]L=n\hbar[/math].This is where the reduced Planck constant [math]\hbar[/math] first appears - in the context of the angular momentum of an electron in the Bohr model of the atom, which is of course, a rotational motion. So it is fairly natural in this case - you have a rotational velocity [math]\omega,[/math] around a circular orbit in the simplest case and to relate this to the period you need the factor [math]2\pi[/math].A photon as it turns out has spin 1, which means that it has an intrinsic angular momentum of exactly [math]\hbar,[/math] that is when projected on an axis.But that was realized only much later on, once the hydrogen atom began to be understood.The spin angular momentum of a photon doesn’t correspond to a rotational velocity in any simple sense.The Bohr quantization condition on the hydrogen atom can be understood in terms of a boundary condition on the so-called de Broglie waves moving around an orbit, which makes the wave single-valued and continuous around the orbit.The Schrödinger equation for the hydrogen atom also leads naturally to the use of [math]\hbar[/math], since the singly periodic elementary functions [math]\sin(\theta)[/math], [math]\sin(\phi)[/math] of the angular coordinates very naturally arise when you consider rotationally invariant systems.So the basic answer is that you need more than just the Einstein-Planck relation [math]E=h\nu[/math] to understand why [math]\hbar[/math] naturally comes in.It really comes in due to the way the theory of quantum mechanics is constructed.Sorry, but it is not possible for me to use simpler vocabulary than that.