Sketch one period of y = 3*sin(2t)

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Answer:

Help with sketching these: Note that these curves, called sinusoidals, take a specific format:

y(x)=Asin(omegax+phi)+y_0

Where `y_0` is the line that the wave oscillates about (called in this case the equillibrium line)

`A` is the amplitude, the distance from the equilibrium line to the maximum or minimum. Amplitudes scales the original wave along the y-axis.

`omega` is the angular frequency. The higher the number, the quicker the wave repeats in the same time (this scales the original wave along the x-axis).

and, `phi` , the phase shift in radians. This moves the graph left or right.

The graph oscillates about the x-axis (`y_0=0`), and has an amplitude of `A=3` and starts how a normal sine wave starts (`phi=0` ).

Since you are asked to sketch one period, you can determine this from the following relationships `T=1/f` and `omega=2pif` This becomes which means that the period for this problem is `T=(2pi)/2=pi` .

Answer: Make your graph of the basic sine for one wave. Now, label the y axis such that the max and min are 3 units away from the x-axis. Then, label the major parts of your wave. The normal sine would start at zero have a max at `pi/2` , zero at `pi` , a min at `(3pi)/2` , and again zero at `2pi` . Since the frequency is twice as much, this happens in half of the angle. Start at zero, max at `pi/4` , zero at `pi/2` , min at `(3pi)/4` and again zero at `pi` .

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