2 we generate the table below 0 4 2 3 4 5 4 3 2 7 4 2

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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...
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