Question

Assertion: In simple harmonic motion, the velocity is maximum when the acceleration is minimum.
Reason: Displacement and velocity of SHM differ in phase by $\dfrac{\pi }{2}$
A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C) Assertion is correct but Reason is incorrect
D) Assertion is incorrect but Reason is correct

Answer 1

Acceleration of a particle executing simple harmonic motion is equal to the second derivative of position of particle with respect to time and is proportional to the position of particle. Velocity of a particle executing simple harmonic motion is equal to the first derivative of position of particle with respect to time.

Complete step-by-step solution:
We are provided with an assertion and a reason. We are required to see if both assertions as well as a reason are correct statements. We are also required to conclude if the given reason is the correct explanation of the given assertion. Assertion states that: In simple harmonic motion, the velocity is maximum when the acceleration is minimum. Reason states that:
Displacement and velocity of SHM differ in phase by $\dfrac{\pi }{2}$. Let us see if both these statements are true.
Firstly, let us consider the position of a particle executing simple harmonic motion. Its wave equation is given by
$x(t)=A\sin \omega t$
where
$x(t)$ is the position of a particle executing simple harmonic motion
$A$ is the maximum amplitude with which the particle oscillates
$\omega$ is the angular frequency
$t$ is the time at which the wave of the particle is recorded
Let this be equation 1.
Now, we know that the velocity of a particle is the first derivative of the position of the particle with respect to time. Taking the first derivative of equation 1, we have
$v(t)={x}'(t)=A\omega \cos \omega t$
where
$v(t)$ is the velocity of the particle
Let this be equation 2.
Moving on, we also know that the acceleration of a particle is the second derivative of the position of the particle. Taking the second derivative of equation 1, we have
$a(t)={{x}'}'(t)=-A{{\omega }^{2}}\sin \omega t$
where
$a(t)$ is the acceleration of the particle
Let this be equation 3.
It is clear from equation 3 that the acceleration of a particle executing simple harmonic motion is proportional to the position of the particle.
From equation 2 and equation 3, it can be concluded that the values of maximum velocity and maximum acceleration are $\omega A$ and ${{\omega }^{2}}A$, respectively. Also, the minimum value of both acceleration and velocity is zero.
Let us understand the above expressions by applying them to the famous spring-mass example of simple harmonic motion. Let a mass be released from a spring of a particular spring constant. We know that the mass executes simple harmonic motion in such a setup. Let us consider two cases, where the mass is at the equilibrium position as well as the extreme position. It is clear that the velocity of the mass is maximum at the equilibrium position because it is this velocity that pushes the mass to the extreme position. The value of the maximum velocity at the equilibrium position can be expressed as
$v(t)={x}'(t)=A\omega \cos \omega t=\omega A$
Clearly,
$\cos \omega t=1\Rightarrow \omega t=0\Rightarrow \sin \omega t=0$
Now, acceleration at the equilibrium position is given by
$a(t)={{x}'}'(t)=-A{{\omega }^{2}}\sin \omega t=0$
Therefore, when the velocity of the mass executing simple harmonic motion is maximum, its acceleration is minimum.
Let us also look into the velocity and acceleration of mass at the extreme position. We know that the velocity of the mass is minimum at the extreme position because the mass comes to a stop at this point due to the force of the string. The minimum velocity of the particle at the extreme position can be expressed as
$v(t)={x}'(t)=A\omega \cos \omega t=0$
Clearly,
$\cos \omega t=0\Rightarrow \omega t=\dfrac{\pi }{2}\Rightarrow \sin \omega t=1$
Now, acceleration of the mass at extreme position is given by
$a(t)={{x}'}'(t)=-A{{\omega }^{2}}\sin \omega t=-{{\omega }^{2}}A$
Therefore, when the velocity of the mass executing simple harmonic motion is minimum, its acceleration is maximum.
Hence, it can be concluded that the given assertion is true.
From equation 1 and equation 2, we can also conclude that position(displacement) of a particle executing simple harmonic motion and its velocity differ by a phase of $\dfrac{\pi }{2}$. Hence, the given reason is also true.
But it cannot be concluded that reason is the correct explanation of the given assertion. Therefore, the correct answer is option B.

Note: Students need to understand that the negative sign in the expression of acceleration of a particle executing simple harmonic motion indicates that acceleration opposes the direction of motion of the particle. For example, in a spring-mass system, the maximum acceleration of mass at the extreme position$(=-{{\omega }^{2}}A)$ is directed towards the equilibrium position and opposes the direction of motion of the mass.