Which of the following will have the highest frequency?

$y=5\mathrm{sin}\left(\frac{x}{2}\right)$

$y=3\mathrm{sin}\left(x\right)$

$y=8\mathrm{sin}\left(2x\right)$

$y=\mathrm{sin}\left(3x\right)$

usagirl007A
2021-05-02
Answered

Which of the following will have the highest frequency?

$y=5\mathrm{sin}\left(\frac{x}{2}\right)$

$y=3\mathrm{sin}\left(x\right)$

$y=8\mathrm{sin}\left(2x\right)$

$y=\mathrm{sin}\left(3x\right)$

You can still ask an expert for help

FieniChoonin

Answered 2021-05-03
Author has **102** answers

A general mathematical wave can be expressed as

$y=A\mathrm{sin}(f\cdot x+\varphi )$

where A is the half-amplitude, f is the frequency, and $\varphi $ is the phase angle.

The half-amplitude is half of the vertical distance from the top of the wave (the crest) to the bottom of the wave (the trough). The frequency (what we're asked about) gives an idea of how "fast" the wave is moving. Graphically, it has to do with how skinny or wide the crests and troughs are - the bigger the frequency, the skinnier the wave and conversely the smaller the frequency, the fatter the wave. And finally the phase angle basically has to do with when exactly you start your stopwatch. Graphically it has to do with where the wave crosses the y-axis.

Let's write your four choices out in this general form.

$y=5\mathrm{sin}\left(\frac{x}{2}\right)\u27f6y=5\mathrm{sin}(\frac{1}{2}x+0)$

$y=3\mathrm{sin}\left(x\right)\u27f6y=3\mathrm{sin}(1x+0)$

$y=8\mathrm{sin}\left(2x\right)\u27f6y=8\mathrm{sin}(2x+0)$

$y=\mathrm{sin}\left(3x\right)\u27f6y=1\mathrm{sin}(3x+0)$

Clearly the last option

$(y=\mathrm{sin}(3x))$

is the one with the highest frequency.

asked 2022-11-05

Solve $\mathrm{sin}(z)=\frac{3+i}{4}$

What i did so far was this:

$$\mathrm{sin}(z)=\frac{3+i}{4}\Rightarrow \frac{{e}^{iz}-{e}^{-iz}}{2i}=\frac{3+i}{4}\Rightarrow {e}^{2iz}-1=\left(\frac{-1+3i}{2}\right){e}^{iz}$$

setting $u={e}^{iz}$ we'll have

$${u}^{2}+\left(\frac{1-3i}{2}\right)u-1=0$$

completing squares

$${(u+\frac{1-3i}{4})}^{2}=\frac{3}{8}(1+i)$$

seting $w=u+\frac{1-3i}{4}$

$${w}^{2}=\frac{3}{8}(1+i)$$

solving

$$w=\pm {2}^{\frac{-5}{4}}\sqrt{3}(\mathrm{cos}\left(\frac{\sqrt{2}}{4}\right)+i\mathrm{sin}\left(\frac{\sqrt{2}}{4}\right))$$

now i just have to substitute this in $w=u+\frac{1-3i}{4}$ and then in $u={e}^{iz}$, but the solution looks really big . So what did i do wrong?

What i did so far was this:

$$\mathrm{sin}(z)=\frac{3+i}{4}\Rightarrow \frac{{e}^{iz}-{e}^{-iz}}{2i}=\frac{3+i}{4}\Rightarrow {e}^{2iz}-1=\left(\frac{-1+3i}{2}\right){e}^{iz}$$

setting $u={e}^{iz}$ we'll have

$${u}^{2}+\left(\frac{1-3i}{2}\right)u-1=0$$

completing squares

$${(u+\frac{1-3i}{4})}^{2}=\frac{3}{8}(1+i)$$

seting $w=u+\frac{1-3i}{4}$

$${w}^{2}=\frac{3}{8}(1+i)$$

solving

$$w=\pm {2}^{\frac{-5}{4}}\sqrt{3}(\mathrm{cos}\left(\frac{\sqrt{2}}{4}\right)+i\mathrm{sin}\left(\frac{\sqrt{2}}{4}\right))$$

now i just have to substitute this in $w=u+\frac{1-3i}{4}$ and then in $u={e}^{iz}$, but the solution looks really big . So what did i do wrong?

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