This question came in the Dhaka University admission exam 2019-20
Q) Two particles are oscillating with simple harmonic motion. If their displacements are described by $x_1=A\sin\omega t$ and $x_2=A\cos\omega t$, what will be the phase difference between them at any instant?
(a) $2\pi$
(b) $\pi$
(c) $\frac{\pi}{2}$
(d) $\frac{\pi}{4}$
My third-party question bank's attempt:
$$x_1=A\sin\omega t$$
$$x_2=A\cos\omega t=A\sin(\omega t\pm\frac{\pi}{2})\tag{1}$$
The phase angle of the first particle is $\omega t$ at time $t$; the phase angle of the second particle is $\omega t\pm \frac{\pi}{2}$ at time $t$.
$\therefore$ Phase difference, $\Delta \delta=(\omega t \pm \frac{\pi}{2})-\omega t=\pm \frac{\pi}{2}$. So, the answer is (c).
My comments:
$(1)$ seems dubious.
$$A\cos\omega t=A\sin(\omega t\pm\frac{\pi}{2})$$
$(1)$ can be broken down to two separate equations $(a)$ and $(b)$:
$$A\cos\omega t=A\sin(\omega t+\frac{\pi}{2})\tag{a}$$
$$A\cos\omega t=A\sin(\omega t-\frac{\pi}{2})\tag{b}$$
$(a)$ is a correct statement. However, $(b)$ seems to be incorrect:
$$A\sin(\omega t-\frac{\pi}{2})$$
$$=-A\sin(\frac{\pi}{2} - \omega t)$$
$$=-A\cos\omega t$$
$$\neq A\cos\omega t$$
So, $(b)$ seems to be an incorrect statement. By extension, (1) also seems to be an incorrect statement.
My questions:
- Isn't the question bank's attempt wrong?
- What corrections have to be made to make the question bank's attempt completely correct?
in $(1)$ would've been sufficient to cover all $\theta$.
– tryingtobeastoic Jul 05 '22 at 14:42