This preview shows page 1. Sign up to view the full content.
Unformatted text preview: rcise 4 from Section 10.5. We ﬁnd
cos(t) + 3 sin(t) = 2 sin t + π so that x(t) = 10e−t/5 sin t + π . Graphing this on the
calculator as y = 10e−x/5 sin x + π reveals some interesting behavior. The sinusoidal nature
continues indeﬁnitely, but it is being attenuated. In the sinusoid A sin(ωx + φ), the coeﬃcient
A of the sine function is the amplitude. In the case of y = 10e−x/5 sin x + π , we can think
of the function A(x) = 10e−x/5 as the amplitude. As x → ∞, 10e−x/5 → 0 which means the
amplitude continues to shrink towards zero. Indeed, if we graph y = ±10e−x/5 along with
y = 10e−x/5 sin x + π , we see this attenuation taking place. This equation corresponds to
the motion of an object on a spring where there is a slight force which acts to ‘damp’, or slow
the motion. An example of this kind of force would be the friction of the object against the
air. In this model, the object oscillates forever, but with smaller and smaller amplitude. y = 10e−x/5 sin x...
View Full Document