Question

A point moves along x-axis according to the equation $x=Asi{n}^2(\omega t-\frac{\pi }{4}).$ Find the amplitude, period and velocity as a function of x.

• A. $\frac{A}{2},\frac{\pi }{\omega },2\omega \sqrt{(A-x)x}$
• B. $3\frac{A}{2},\frac{\pi }{\omega },\omega \sqrt{(A-x)x}$
• C. $\frac{3A}{2},\frac{\pi }{\omega },3\omega \sqrt{(A-x)x}$
• D. $\frac{A}{2},2\frac{\pi }{\omega },2\omega \sqrt{(A-x)x}$

Answer 1

The correct option is A $\frac{A}{2},\frac{\pi }{\omega },2\omega \sqrt{(A-x)x}$

$x=Asi{n}^2(\omega t-\frac{\pi }{4})$
$=A(\sin \omega t\cos \frac{\pi }{4}-\cos \omega t\sin \frac{\pi }{4}{)}^2=A\frac{1}{2}(\sin \omega t-\cos \omega t{)}^2$
$=\frac{A}{2}(1-\sin 2\omega t)....\left(1\right)$
Maximum displacement occurs when $\sin 2\omega t=0$
$\therefore$ Amplitude = $\frac{A}{2}$
Period is given by $2\omega =2\omega =\frac{2\pi }{T}\therefore T=\frac{\pi }{\omega }$
$\frac{dx}{dt}=\frac{A}{2}\left(2\omega \cos 2\omega t\right)=A\omega \cos 2\omega t$
$=A\omega \sqrt{1-{\sin }^{2}2\omega t}$
$=A\omega \sqrt{1-{(1-\frac{2x}{A})}^2}$ (because from (1), ) $1-\frac{2x}{A}=\sin 2\omega t$
$=A\omega \sqrt{(2-\frac{2x}{A})\left(\frac{2x}{A}\right)}=$
$2A\omega \sqrt{(1-\frac{x}{A})\left(\frac{x}{A}\right)}=2\omega \sqrt{(A-x)x}$