Question

Simple harmonic motions are special cases of periodic motion in which the moving object feels:
A. acceleration in the opposite direction to its displacement
B. acceleration in the same direction to its displacement
C. velocity in the opposite direction to its displacement
D. velocity in the same direction to its displacement

Answer 1

Start by assuming the time, when the object will be at the mean position .Then using the SHM equation for displacement of object $x=A\sin(\omega t)$, when differentiated, gives the velocity and the acceleration of the motion.

Formula used:
Also, $v=\dfrac{displacement}{time}$ Or $v=\dfrac{dx}{dt}$ and $a=\dfrac{velocity}{time}$.
Or$a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$

Complete answer:
SHM or simple harmonic motion is the motion caused due to the restoring force; it is directly proportional to the displacement of the object from its mean position. And it is always directed towards the mean.
Let us assume that the displacement of object is given as $x=A\sin(\omega t)$
We know that the velocity of the object is given as $v=\dfrac{displacement}{time}$ Or $v=\dfrac{dx}{dt}$ and the acceleration of the object is given as $a=\dfrac{velocity}{time}$ Or $a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$
Then the velocity of the given object is $v=\dfrac{dx}{dt}=\dfrac{A sin(\omega t)}{dt}=Acos(\omega t)$
Clearly, the direction of the velocity doesn’t change with respect to the displacement. Hence option D is correct.
Also, the acceleration of the object is given by, $a=\dfrac{dv}{dt}=\dfrac{Acos(\omega t)}{dt}=-A sin (\omega t)$
Here, the negative sign implies that the direction of acceleration is opposite to that of the direction of displacement.

So, the correct answer is “Option A and D”.

Note:
Remember SHM motions are sinusoidal in nature. Also see that $\dfrac{dx}{dt}=\dfrac{1}{\dfrac{dt}{dx}}$, this is the most important step in this question. Also note that$v=\dfrac{displacement}{time}$ Or $v=\dfrac{dx}{dt}$ and $a=\dfrac{velocity}{time}$.Or $a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$. To calculate, $a$ we must differentiate only $v$ with respect to $t$ and not $\dfrac{dt}{dx}$. Also, to solve this sum one needs to basic differentiation of trigonometric identities.