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15 • Oscillatory Motion - ECHSPhysics

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O<br />

O<br />

O<br />

x<br />

v<br />

a<br />

T<br />

2<br />

T<br />

2<br />

T<br />

2<br />

(a)<br />

T<br />

T<br />

T<br />

3T<br />

2<br />

3T<br />

2<br />

3T<br />

2<br />

Figure <strong>15</strong>.8 (a) Position, velocity, and acceleration versus time for a block undergoing<br />

simple harmonic motion under the initial conditions that at t 0, x(0) A and v(0) 0.<br />

(b) Position, velocity, and acceleration versus time for a block undergoing simple harmonic<br />

motion under the initial conditions that at t 0, x(0) 0 and v(0) v i.<br />

our solutions for x(t) and v(t) (Eqs. <strong>15</strong>.6 and <strong>15</strong>.<strong>15</strong>) obey the initial conditions that<br />

x(0) A and v(0) 0:<br />

x(0) A cos A<br />

v(0) A sin 0<br />

These conditions are met if we choose 0, giving x A cos t as our solution. To<br />

check this solution, note that it satisfies the condition that x(0) A, because<br />

cos 0 1.<br />

The position, velocity, and acceleration versus time are plotted in Figure <strong>15</strong>.8a for<br />

this special case. The acceleration reaches extreme values of 2 A when the position<br />

has extreme values of A. Furthermore, the velocity has extreme values of A,<br />

which both occur at x 0. Hence, the quantitative solution agrees with our qualitative<br />

description of this system.<br />

Let us consider another possibility. Suppose that the system is oscillating and we<br />

define t 0 as the instant that the particle passes through the unstretched position<br />

of the spring while moving to the right (Fig. <strong>15</strong>.9). In this case we must require that<br />

our solutions for x(t) and v(t) obey the initial conditions that x(0) 0 and<br />

v(0) v i :<br />

x(0) A cos 0<br />

v(0) A sin v i<br />

The first of these conditions tells us that /2. With these choices for ,<br />

the second condition tells us that A v i/. Because the initial velocity is positive<br />

and the amplitude must be positive, we must have /2. Hence, the solution is<br />

given by<br />

x v i<br />

t<br />

t<br />

t<br />

cos <br />

<br />

t <br />

2 <br />

The graphs of position, velocity, and acceleration versus time for this choice of<br />

t 0 are shown in Figure <strong>15</strong>.8b. Note that these curves are the same as those<br />

in Figure <strong>15</strong>.8a, but shifted to the right by one fourth of a cycle. This is described<br />

mathematically by the phase constant /2, which is one fourth of a full cycle<br />

of 2.<br />

O<br />

O<br />

O<br />

x<br />

v<br />

a<br />

T<br />

2<br />

T<br />

2<br />

T<br />

2<br />

SECTION <strong>15</strong>.2 <strong>•</strong> Mathematical Representation of Simple Harmonic <strong>Motion</strong> 459<br />

T<br />

T<br />

T<br />

(b)<br />

3T<br />

2<br />

3T<br />

2<br />

3T<br />

2<br />

t<br />

t<br />

t<br />

x i = 0<br />

t = 0<br />

v = v i<br />

x = 0<br />

Active Figure <strong>15</strong>.9 The<br />

block–spring system is undergoing<br />

oscillation, and t 0 is defined at<br />

an instant when the block passes<br />

through the equilibrium position<br />

x 0 and is moving to the right<br />

with speed v i.<br />

At the Active Figures link<br />

at http://www.pse6.com, you<br />

can compare the oscillations of<br />

two blocks with different<br />

velocities at t 0 to see that<br />

the frequency is independent of<br />

the amplitude.<br />

m<br />

v i

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