Understanding the Phase Difference Between Velocity and Acceleration in Simple Harmonic Motion (SHM)

This explanation details the relationship between the instantaneous velocity and instantaneous acceleration of a particle undergoing simple harmonic motion (SHM). We will determine the specific phase difference between these two quantities.

What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) describes an oscillatory motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction. Key characteristics include amplitude, angular frequency, and phase.

Mathematical Description of SHM Variables

Let's represent the displacement, instantaneous velocity, and instantaneous acceleration of a particle in SHM using standard equations. We can start with the displacement equation. For simplicity, let's assume the motion starts from the maximum displacement position at time $t=0$. The displacement ($x$) as a function of time ($t$) can be written as:

$x(t) = A \cos(\omega t)$

Here:

• $A$ represents the amplitude (maximum displacement).
• $\omega$ represents the angular frequency.
• $t$ represents time.

The instantaneous velocity ($v$) is the first derivative of displacement with respect to time:

$v(t) = \frac{dx}{dt} = \frac{d}{dt} [A \cos(\omega t)]$

Calculating the derivative:

$v(t) = -A\omega \sin(\omega t)$

To make the phase comparison easier, we can express the velocity using a cosine function. Using the trigonometric identity $-\sin(\theta) = \cos(\theta + \frac{\pi}{2})$, we get:

$v(t) = A\omega \cos(\omega t + \frac{\pi}{2})$

From this, we can see the phase of the instantaneous velocity is $(\omega t + \frac{\pi}{2})$.

The instantaneous acceleration ($a$) is the first derivative of velocity with respect to time (or the second derivative of displacement):

$a(t) = \frac{dv}{dt} = \frac{d}{dt} [-A\omega \sin(\omega t)]$

Calculating the derivative:

$a(t) = -A\omega^2 \cos(\omega t)$

Similarly, we express acceleration using a cosine function. Using the trigonometric identity $-\cos(\theta) = \cos(\theta + \pi)$, we get:

$a(t) = A\omega^2 \cos(\omega t + \pi)$

From this, the phase of the instantaneous acceleration is $(\omega t + \pi)$.

Determining the Phase Difference

Now, we compare the phases of the instantaneous velocity and instantaneous acceleration.

• Phase of Instantaneous Velocity = $\omega t + \frac{\pi}{2}$
• Phase of Instantaneous Acceleration = $\omega t + \pi$

The phase difference ($\Delta \phi$) is the difference between these two phases:

$\Delta \phi = \text{Phase}(\text{acceleration}) - \text{Phase}(\text{velocity})$

$\Delta \phi = (\omega t + \pi) - (\omega t + \frac{\pi}{2})$

$\Delta \phi = \pi - \frac{\pi}{2}$

$\Delta \phi = \frac{\pi}{2} \text{ rad}$

Alternatively, we could calculate it as Phase(velocity) - Phase(acceleration), which yields $-\frac{\pi}{2}$ rad. The magnitude of the phase difference remains $\frac{\pi}{2}$ radians.

Conclusion

The phase difference between the instantaneous velocity and the instantaneous acceleration in simple harmonic motion is $\frac{\pi}{2}$ radians. This signifies that velocity and acceleration are always out of phase by a quarter of a complete oscillation cycle. For instance, when the velocity is at its maximum magnitude (either positive or negative), the acceleration is zero, and when the acceleration is at its maximum magnitude, the velocity is zero.