In general, you can talk about any function as a transformation of a parent function. The parent function for all sine functions is (or whatever you want to call the variables - I've used d for displacement and t for time). This parent function has an amplitude of 1 and a period of .

A transformation would look like this: , where A is the amplitude (how much the graph is stretched or squshed vertically), b is a factor related to the frequency or period (it squishes or stretches the curve in a horizontal direction on the graph), c is the phase shift (which effects where the graph intersects the vertical axis - when t=0) and u is the vertical shift.

In your case, the amplitude is 1.6.

Since f=20 hertz, a complete cycle takes 1/20 of a second. Since the sine function has a natural period of , it would need to be squished by a factor of (so this is 'b' in my version of the formula the formula).

Regarding the phase shift, this is the amount the parent graph should be moved to meet the initial conditions. So, . Solving this equation gives . In your equation, you have distributed the through the , so this is the equivalent of what you got as well. However, from what you describe there is no way of knowing whether the displacement is increasing or decreasing at time t=0. It could be doing either. Your equation is correct if the displacement is increasing. But if the displacement is decreasing when t=0, then you would have to move the graph further. The actual phase shift of your solution is . If the displacement were decreasing, the displacement would be half a cycle minus this amount, or . Multiplying to get this in the form or your equation would give you a phase angle of .

So there are two equations that will satisfy these conditions:

and

.

Overall a good job by you!!!

If the formula for 'lead time' is , then you should just plug in those numbers. This is the equivalent of the 'phase shift' in my formula: or . Either way, it is 0.002529 seconds away from its center of displacement (frequently called the equilibrium position), but it would pass this position before t=0 (and headed in the positive direction) in the first case and after t=0 (and heading in the negative direction) in the second case.