I have the question "An oscillator undergoes SHM (simple harmonic motion) with a frequency of 7.2 s^-1. If it's velocity is 12 ms^-1 at a displacement of 40 cm, what is the maximum amplitude ?"
Here is my attempt. Is this correct ?
So I rearranged the velocity formula to make the amplitude(A) the subject. If my answer is correct are my units also correct ?
On
$$v=\pm\omega\sqrt{A^2-x^2}\longrightarrow{v\over\omega}=\pm\sqrt{A^2-x^2}$$ $${v^2\over\omega^2}=A^2-x^2$$ $$A=\pm\sqrt{{v^2\over\omega^2}+x^2}$$
We can take $|A|$, so:
$$|A|=\sqrt{{v^2\over\omega^2}+x^2}=\sqrt{\left({v\over2\pi\nu}\right)^2+x^2}=\sqrt{\left({12m/s\over2\pi 7.2 s^{-1}}\right)^2+0,40^2m^2}$$
and the answer is: $A=0,479960345m\approx0,48 m$, thus the unit is meter.
In 6th line of your attempt you have made a mistake. You can't cancel $\omega$ with a single term from numerator. At best you can do:
$$ A = \sqrt{\frac{v^2}{\omega^2} + x^2} $$ Or $$ A = \frac{\sqrt{v^2+ \omega^2 \times x^2}}{\omega} $$
Your unit is incorrect because of this error and your addition of $v^2$ and $\omega x^2$ is dimensionally inconsistent.
Lines 2 and 3: Remove the ± sign from there. You squared both sides and square of real number is always positive.