**Unformatted text preview: **ing.16 y = 2(x + 3) sin 2x + π
4 y = 2(x + 3) sin 2x + π
4 , y = ±2(x + 3) 3. Last, but not least, we come to x(t) = 5 sin(6t) − 5 sin(8t). To ﬁnd the period of this function,
we need to determine the length of the smallest interval on which both f (t) = 5 sin(6t) and
g (t) = 5 sin(8t) complete a whole number of cycles. To do this, we take the ratio of their
6
frequencies and reduce to lowest terms: 8 = 3 . This tells us that for every 3 cycles f makes,
4
g makes 4. In other words, the period of x(t) is three times the period of f (t) (which is
four times the period of g (t)), or π . We graph y = 5 sin(6x) − 5 sin(8x) over [0, π ] on the
calculator to check this. This equation of motion also results from ‘forced’ motion, but here
the frequency of the external oscillation is diﬀerent than that of the object on the spring.
Since the sinusoids here have diﬀerent frequencies, they are ‘out of sync’ and do not amplify
each other as in the previous example. Taking things a step further, we can use a sum to
product identity to rewrite x(t) = 5 sin(6t) − 5 sin(8t) as x(t) = −10 sin(t) cos(7t). The lowe...

View
Full
Document