Simple Harmonic Motion: Equations of Motion

1. What is Simple Harmonic Motion?

At its heart, Simple Harmonic Motion is the motion of an object where the restoring force is directly proportional to the object's displacement from its equilibrium (or mean) position and is always directed towards that position.

Mathematically, this is expressed by Hooke's Law:
$F=-kx$
where:

• $F$ is the restoring force.
• $x$ is the displacement from the equilibrium position.
• $k$ is a positive constant called the force constant or spring constant.
• The negative sign indicates that the force is always opposite to the direction of displacement.

A classic example is a mass attached to a spring oscillating on a frictionless surface.

2. Visualizing SHM: The Connection to Circular Motion

Before diving into the equations, it's incredibly helpful to visualize where they come from. A powerful way to do this is to see SHM as the projection of Uniform Circular Motion (UCM) onto a diameter.

Imagine a particle moving in a circle at a constant speed. If you track the shadow (projection) of this particle on one of the diameters of the circle, you'll see that the shadow moves back and forth with Simple Harmonic Motion.

To see this in action, please watch the following video. It provides an excellent visual demonstration of how the sinusoidal graphs of SHM emerge from circular motion.

3. Deriving the Mathematical Model of SHM

Now, let's formalize this visual understanding into mathematical equations. This is a critical step for problem-solving.

Displacement $x(t)$

As you saw in the video, the projection of a particle in UCM follows a sinusoidal path. The general equation for the displacement $x$ of a particle in SHM as a function of time $t$ is:
$x(t)=A\cos (\omega t+\varphi )$
Let's break down each term:

• $A$: Amplitude. This is the maximum displacement from the mean position. It's the radius of the reference circle.
• $\omega$: Angular Frequency. Measured in radians per second (rad/s), it tells us how quickly the oscillations occur. It's related to the period $T$ (time for one full oscillation) and frequency $f$ (oscillations per second, in Hz) by:
  $\omega =\frac{2\pi }{T}=2\pi f$
• $(\omega t+\varphi )$: Phase. This is the argument of the cosine function. It determines the state of the oscillation at any given time $t$.
• $\varphi$: Phase Constant (or phase angle). This value tells us the position of the particle at $t=0$. It is determined by the initial conditions of the motion.

Note: You might also see the equation written as $x(t)=A\sin (\omega t+\varphi )$. Both are equally valid; the choice simply depends on where you define the starting point ($t=0$) of the motion. The cosine function is convenient when the motion starts from an extreme position.

To solidify your understanding of these derivations, please read through the following resource. It is from Aakash, a source you might find familiar, and covers these derivations in a style relevant to your NEET preparation.

Velocity $v(t)$

Since velocity is the rate of change of displacement, we can find the velocity equation by differentiating the displacement equation with respect to time. Given your background in 12th-grade science, this application of calculus should be familiar.

If $x(t)=A\cos (\omega t+\varphi )$, then:
$v(t)=\frac{dx}{dt}=\frac{d}{dt}[A\cos (\omega t+\varphi )]$
Using the chain rule, we get:
$v(t)=-A\omega \sin (\omega t+\varphi )$

The maximum velocity, or velocity amplitude, is $v_{max}=A\omega$. This occurs when $\sin (\omega t+\varphi )=\pm 1$, which corresponds to the particle passing through the mean position ($x=0$).

Acceleration $a(t)$

Similarly, acceleration is the rate of change of velocity. We differentiate the velocity equation with respect to time:
$a(t)=\frac{dv}{dt}=\frac{d}{dt}[-A\omega \sin (\omega t+\varphi )]$
$a(t)=-A{\omega }^2\cos (\omega t+\varphi )$

The maximum acceleration, or acceleration amplitude, is $a_{max}=A{\omega }^2$. This occurs when $\cos (\omega t+\varphi )=\pm 1$, which corresponds to the particle being at its extreme positions ($x=\pm A$).

Notice something interesting by substituting $x(t)=A\cos (\omega t+\varphi )$ into the acceleration equation:
$a(t)=-{\omega }^{2}x(t)$
This is the hallmark equation of SHM. It shows that acceleration is directly proportional to displacement and in the opposite direction, which is the fundamental definition we started with ($F=ma⟹a=F/m=-(k/m)x$). This also gives us a very important relation: ${\omega }^2=k/m$.

For a quick summary of these derivations, you can watch this short video.

4. Phase Relationships and Graphs

The equations for $x(t)$, $v(t)$, and $a(t)$ are all sinusoidal, but they are shifted in phase relative to each other.

• Velocity $v(t)$ leads displacement $x(t)$ by $\pi /2$ radians (90°).
• Acceleration $a(t)$ leads velocity $v(t)$ by $\pi /2$ radians (90°).
• Acceleration $a(t)$ leads displacement $x(t)$ by $\pi$ radians (180°), meaning they are always in opposite directions (out of phase).

The image below summarizes these relationships beautifully.

The relationships between the slopes of these graphs are also a direct consequence of calculus.

5. Mathematical Problem-Solving

Your goal is to prepare for NEET, so let's apply these concepts to solve a typical problem. We will work through it step-by-step.

Here is a comprehensive problem from a university physics course that covers all the aspects we've discussed.

Let's break down the solution to Question 1:

Problem Statement: A 1.75-kg particle moves as a function of time as follows:
$x=4\cos (1.33t+\pi /5)$

(a) Find amplitude, frequency, angular frequency, and period.

• By comparing the given equation to the general form $x=A\cos (\omega t+\varphi )$, you can directly identify the values:
  • Amplitude $A=4$ m
  • Angular Frequency $\omega =1.33$ rad/s
  • Phase Constant $\varphi =\pi /5$ rad
• From $\omega$, you can calculate the period $T$ and frequency $f$:
  • $T=2\pi /\omega =2\pi /1.33\approx 4.72$ s
  • $f=1/T\approx 0.21$ Hz

(b) Find the equation for velocity.

• Use the velocity formula we derived: $v=-A\omega \sin (\omega t+\varphi )$
• Substitute the known values:
  • $v=-(4)(1.33)\sin (1.33t+\pi /5)$
  • $v=-5.32\sin (1.33t+\pi /5)$

(c) Find the equation for acceleration.

• Use the acceleration formula: $a=-A{\omega }^2\cos (\omega t+\varphi )$ or $a=-{\omega }^{2}x$
• Substitute the known values:
  • $a=-(1.33{)}^2[4\cos (1.33t+\pi /5)]$
  • $a=-7.08\cos (1.33t+\pi /5)$

The resource Physics 1120: Simple Harmonic Motion Solutions also provides an alternative, quicker method using the reference circle for solving time-dependent problems (part e of Question 1), which can be a very useful technique for competitive exams. The resource from Aakash, Displacement, Velocity & Acceleration of a particle in SHM, also contains several solved practice problems that I encourage you to work through to build your speed and accuracy.

Conclusion

In this lesson, we have laid the complete mathematical foundation for Simple Harmonic Motion.

Key Takeaways:

• SHM is defined by the relation $F=-kx$, leading to the characteristic equation $a=-{\omega }^{2}x$.
• We can model SHM using sinusoidal functions. The displacement of a particle is given by $x(t)=A\cos (\omega t+\varphi )$.
• Velocity and acceleration are found by taking successive time derivatives of the displacement:
  • $v(t)=-A\omega \sin (\omega t+\varphi )$
  • $a(t)=-A{\omega }^2\cos (\omega t+\varphi )$
• Velocity is maximum at the mean position ($x=0$) and zero at the extremes.
• Acceleration is maximum at the extreme positions ($x=\pm A$) and zero at the mean position.

In our next lesson, we will build upon this foundation to calculate the kinetic, potential, and total energy in oscillating systems. Understanding the energy transformations during SHM is the next logical step and a frequent topic in exam questions.