How to differenciate Asinwt

Its Aw coswt
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we shall find the period of oscillation T of the pendulum which consists of the ball of mass m and the spring of stiffness k. If the ball is shifted from the initial point (where the spring is not deformed) by the distance x, it is acted upon by the force -kx from the side of the spring. Additionally, the constant gravitational force mg is applied to the ball. On the other hand, according to the second Newton's law the sum of all forces applied to a body is equal to ma, where a is the acceleration. So we can write the differential equation for the oscillating pendulum:

md2x/dt2= -kx + mg

g is the acceleration of the free fall in the gravitational field, d2x/dt2 is the second derivative of coordinate x with respect to time t. The solution of this equation follows:

x  = Asin[(k/m)1/2t + j] + mg/k

We can see from this formula that the period of oscillation is equal to

T = 2p(m/k)1/2

and the angular frequency w is equal to

w = (k/m)1/2

The amplitude of oscillation A and the phase of oscillation j depend upon the initial conditions (at the moment t=0): initial deviation of the ball x0 and the initial velocity v0. At the point of equilibrium the spring is extended by the distance mg/k.

Let us imagine that the oscillating ball is connected to a pen which draws a line on a moving paper ribbon. If the ribbon is moving evenly in the horizontal direction, then the pen will draw a sinusoid on it. If we know the velocity of the ribbon’s motion and the period of the sinusoid, we can calculate the period of oscillation.

Generally, the oscillator is acted upon by the force of friction proportional to the velocity v of motion: F=av. In the case of the ball on the spring, this force appears because of the air friction and internal friction in the material of the spring. As a result, the amplitude of oscillation will attenuate with time. The equation of the free harmonic oscillator with attenuation can be written as:

m(d2x/dt2) + a (dx/dt) + kx = mg

where a is the coefficient of friction. This equation can be rewritten as

d2x/dt2 + 2g(dx/dt) + W2x = g

where 2g = a/m; W2=k/m

In the case when W2 > g2 the solution of this equation is as follows:

x = Ae-gtcos(wt + j )

The period of oscillation depends in this case upon the attenuation coefficient g :

T = 2p/w = 2p/(W2 - g2)1/2
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Tha tray agen
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