Welcome, future engineers! Today, we're going to take a deep dive into the fascinating world of
Simple Harmonic Motion (SHM). This isn't just about memorizing equations; it's about understanding the underlying physics, the rhythm, and the energy transformations that make SHM so fundamental in mechanics. SHM is a cornerstone concept for JEE, often appearing in both Mains and Advanced, so let's build a rock-solid understanding.
---
###
1. Unveiling Simple Harmonic Motion: The Defining Equation
At its heart, SHM is a special type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction to the displacement. Think of it like a strict manager always pulling you back to the center!
Let's formalize this. If 'x' is the displacement from the equilibrium position, the restoring force 'F' is given by:
$$F = -kx$$
Here, 'k' is the
spring constant (or force constant), a measure of the stiffness of the system. The negative sign is crucial: it indicates that the force always acts towards the equilibrium position. If you displace the object to the right (positive x), the force pulls it to the left (negative F), and vice-versa.
Now, recall Newton's Second Law of Motion: $F = ma$. Substituting the restoring force:
$$ma = -kx$$
Since acceleration $a = frac{d^2x}{dt^2}$ (the second derivative of displacement with respect to time), we get:
$$mfrac{d^2x}{dt^2} = -kx$$
Rearranging this equation, we arrive at the
differential equation of Simple Harmonic Motion:
$$frac{d^2x}{dt^2} + frac{k}{m}x = 0$$
This equation is the mathematical fingerprint of SHM. Any system whose motion can be described by this equation will execute SHM.
To simplify, we define $omega^2 = frac{k}{m}$, where $omega$ is the
angular frequency of oscillation. This gives us:
$$frac{d^2x}{dt^2} + omega^2x = 0$$
JEE Focus: Understanding this differential equation is key. Sometimes, problems might present a system and ask you to show it undergoes SHM, which means deriving this form.
---
###
2. The General Solution: Position, Velocity, and Acceleration in SHM
The differential equation $frac{d^2x}{dt^2} + omega^2x = 0$ is a second-order linear differential equation. Its general solution describes the position of the oscillating particle as a function of time. The most common forms are sinusoidal:
- $$x(t) = Asin(omega t + phi)$$
- $$x(t) = Acos(omega t + phi)$$
Both are equivalent, differing only in the phase constant. Let's use $x(t) = Asin(omega t + phi)$ for our derivations.
*
A: Amplitude
This is the maximum displacement from the equilibrium position. It's always a positive quantity. The particle oscillates between $+A$ and $-A$.
*
$omega$: Angular Frequency
Measured in radians per second (rad/s). We know $omega = sqrt{frac{k}{m}}$. It's related to the frequency (f) and time period (T) of oscillation:
$$omega = 2pi f = frac{2pi}{T}$$
*
Frequency (f): Number of oscillations per second (Hz).
*
Time Period (T): Time taken for one complete oscillation (s).
*
$(omega t + phi)$: Phase
This entire term represents the "state" of oscillation at time 't'. It tells us where the particle is in its cycle.
*
$phi$: Initial Phase / Phase Constant
This constant is determined by the initial conditions (position and velocity) of the particle at $t=0$. It essentially tells us where the oscillation "starts" in its cycle.
#### 2.1 Velocity in SHM
To find the velocity $v(t)$, we differentiate the displacement $x(t)$ with respect to time:
$$v(t) = frac{dx}{dt} = frac{d}{dt}[Asin(omega t + phi)]$$
$$v(t) = Aomegacos(omega t + phi)$$
Key Insight:
* The velocity is maximum when $cos(omega t + phi) = pm 1$. So,
$v_{max} = pm Aomega$. This occurs when the particle is passing through the equilibrium position ($x=0$).
* The velocity is zero at the extreme positions ($x = pm A$), where $sin(omega t + phi) = pm 1$ and thus $cos(omega t + phi) = 0$.
#### 2.2 Acceleration in SHM
To find the acceleration $a(t)$, we differentiate the velocity $v(t)$ with respect to time:
$$a(t) = frac{dv}{dt} = frac{d}{dt}[Aomegacos(omega t + phi)]$$
$$a(t) = -Aomega^2sin(omega t + phi)$$
We can also express acceleration in terms of displacement:
$$a(t) = -omega^2 [Asin(omega t + phi)]$$
$$a(t) = -omega^2 x(t)$$
This confirms our original differential equation $a = -omega^2 x$.
Key Insight:
* The acceleration is maximum when $sin(omega t + phi) = pm 1$. So,
$a_{max} = mp Aomega^2$. This occurs at the extreme positions ($x=pm A$).
* The acceleration is zero at the equilibrium position ($x=0$), where $sin(omega t + phi) = 0$. This makes sense, as the net force is zero at equilibrium.
Quantity |
Equation |
Value at Equilibrium ($x=0$) |
Value at Extremes ($x=pm A$) |
|---|
Displacement ($x$) |
$Asin(omega t + phi)$ |
$0$ |
$pm A$ |
Velocity ($v$) |
$Aomegacos(omega t + phi)$ |
$pm Aomega$ (Maximum) |
$0$ (Minimum) |
Acceleration ($a$) |
$-Aomega^2sin(omega t + phi) = -omega^2 x$ |
$0$ (Minimum) |
$mp Aomega^2$ (Maximum) |
---
###
3. Energy in Simple Harmonic Motion
One of the most beautiful aspects of SHM is the continuous interconversion between kinetic and potential energy, while the total mechanical energy remains conserved (in an ideal, undamped system).
#### 3.1 Kinetic Energy (KE)
The kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by $KE = frac{1}{2}mv^2$.
Substituting $v(t) = Aomegacos(omega t + phi)$:
$$KE(t) = frac{1}{2}m[Aomegacos(omega t + phi)]^2$$
$$KE(t) = frac{1}{2}mA^2omega^2cos^2(omega t + phi)$$
Since $omega^2 = frac{k}{m}$, we have $momega^2 = k$. Substituting this:
$$KE(t) = frac{1}{2}kA^2cos^2(omega t + phi)$$
Important Points:
* KE is maximum at the equilibrium position ($x=0$), where $v$ is maximum. $KE_{max} = frac{1}{2}kA^2$.
* KE is zero at the extreme positions ($x=pm A$), where $v$ is zero.
#### 3.2 Potential Energy (PE)
For a spring-mass system, the potential energy stored when the spring is stretched or compressed by 'x' is $PE = frac{1}{2}kx^2$.
Substituting $x(t) = Asin(omega t + phi)$:
$$PE(t) = frac{1}{2}k[Asin(omega t + phi)]^2$$
$$PE(t) = frac{1}{2}kA^2sin^2(omega t + phi)$$
Important Points:
* PE is maximum at the extreme positions ($x=pm A$), where $x$ is maximum. $PE_{max} = frac{1}{2}kA^2$.
* PE is zero at the equilibrium position ($x=0$), where $x$ is zero.
#### 3.3 Total Mechanical Energy (TME)
The total mechanical energy is the sum of kinetic and potential energy:
$$TME(t) = KE(t) + PE(t)$$
$$TME(t) = frac{1}{2}kA^2cos^2(omega t + phi) + frac{1}{2}kA^2sin^2(omega t + phi)$$
$$TME(t) = frac{1}{2}kA^2[cos^2(omega t + phi) + sin^2(omega t + phi)]$$
Since $cos^2 heta + sin^2 heta = 1$:
$$TME = frac{1}{2}kA^2$$
Crucial Conclusion: The total mechanical energy in SHM is constant and proportional to the square of the amplitude. It does not depend on time! This confirms the principle of conservation of energy for an ideal SHM system.
We can also write this as $TME = frac{1}{2}momega^2A^2$.
Graphical Representation of Energy:
If you plot KE, PE, and TME against displacement 'x' (ranging from $-A$ to $+A$):
*
KE: A downward parabola, maximum at $x=0$, zero at $x=pm A$. ($KE = frac{1}{2}k(A^2 - x^2)$)
*
PE: An upward parabola, zero at $x=0$, maximum at $x=pm A$. ($PE = frac{1}{2}kx^2$)
*
TME: A horizontal line at $frac{1}{2}kA^2$.
This graph beautifully illustrates the continuous transformation of energy. As the particle moves from $x=0$ to $x=A$, KE decreases while PE increases, with their sum remaining constant.
#### 3.4 Average Kinetic and Potential Energy
Over one complete cycle of oscillation, the average kinetic energy and average potential energy are equal.
Average value of $sin^2( heta)$ or $cos^2( heta)$ over a full cycle is $frac{1}{2}$.
Therefore:
$$langle KE
angle = frac{1}{2}kA^2 langle cos^2(omega t + phi)
angle = frac{1}{2}kA^2 imes frac{1}{2} = frac{1}{4}kA^2$$
$$langle PE
angle = frac{1}{2}kA^2 langle sin^2(omega t + phi)
angle = frac{1}{2}kA^2 imes frac{1}{2} = frac{1}{4}kA^2$$
Important Takeaway:
Over one full cycle, $langle KE
angle = langle PE
angle = frac{1}{4}kA^2 = frac{1}{2}TME$. This is a frequently tested concept in JEE.
---
###
4. Phase Relationships & Advanced Insights
*
Phase Difference:
* Velocity leads displacement by $frac{pi}{2}$ radians (or $90^circ$). When displacement is maximum, velocity is zero; when displacement is zero, velocity is maximum.
* Acceleration leads velocity by $frac{pi}{2}$ radians, and leads displacement by $pi$ radians (or $180^circ$). When displacement is positive maximum, acceleration is negative maximum. They are always in opposite directions.
*
Conditions for SHM:
1. The restoring force must be directly proportional to the displacement from equilibrium ($F propto -x$).
2. The system must be conservative (no energy dissipation like friction or air resistance) for the total mechanical energy to remain constant.
*
Example: Simple Pendulum (Small Oscillations)
For a simple pendulum of length 'L' and mass 'm', the restoring force component (tangential) for small angular displacement $ heta$ is $F = -mgsin heta approx -mg heta$.
Since $x = L heta$, we have $ heta = x/L$.
So, $F approx -mg(x/L) = -left(frac{mg}{L}
ight)x$.
Comparing this with $F = -kx$, we get $k = frac{mg}{L}$.
Therefore, the angular frequency is $omega = sqrt{frac{k}{m}} = sqrt{frac{mg/L}{m}} = sqrt{frac{g}{L}}$.
And the time period is $T = frac{2pi}{omega} = 2pisqrt{frac{L}{g}}$.
This shows how a simple pendulum *approximates* SHM for small angles. This derivation is critical for JEE.
---
###
5. Solved Example: Applying SHM Equations and Energy
Problem: A particle executes SHM with an amplitude of 10 cm and a time period of 2 s. Assume it starts from its positive extreme position at $t=0$. Calculate:
a) The angular frequency ($omega$).
b) The equation for displacement $x(t)$.
c) The velocity and acceleration at $t = 0.5$ s.
d) The kinetic, potential, and total energy at $t = 0.5$ s, if the mass of the particle is 100 g.
Solution:
Given: $A = 10 ext{ cm} = 0.1 ext{ m}$, $T = 2 ext{ s}$, $m = 100 ext{ g} = 0.1 ext{ kg}$.
Initial condition: At $t=0$, $x = +A$ (positive extreme position).
a) Angular frequency ($omega$):
$$omega = frac{2pi}{T} = frac{2pi}{2} = pi ext{ rad/s}$$
b) Equation for displacement $x(t)$:
Since the particle starts from the positive extreme position ($x=+A$) at $t=0$, we choose the cosine function for displacement:
$x(t) = Acos(omega t + phi)$.
At $t=0$, $x(0) = Acos(phi) = A$.
This implies $cos(phi) = 1$, so $phi = 0$.
Therefore, the equation for displacement is:
$$x(t) = Acos(omega t) = 0.1cos(pi t) ext{ m}$$
c) Velocity and acceleration at $t = 0.5$ s:
First, find the general equations for velocity and acceleration:
$$v(t) = frac{dx}{dt} = frac{d}{dt}[0.1cos(pi t)] = -0.1pisin(pi t) ext{ m/s}$$
$$a(t) = frac{dv}{dt} = frac{d}{dt}[-0.1pisin(pi t)] = -0.1pi^2cos(pi t) ext{ m/s}^2$$
Now, substitute $t = 0.5$ s:
$$x(0.5) = 0.1cos(pi imes 0.5) = 0.1cos(pi/2) = 0.1 imes 0 = 0 ext{ m}$$
(This means the particle is at the equilibrium position at $t=0.5$ s).
$$v(0.5) = -0.1pisin(pi imes 0.5) = -0.1pisin(pi/2) = -0.1pi imes 1 = -0.1pi ext{ m/s}$$
$$v(0.5) approx -0.1 imes 3.14 = -0.314 ext{ m/s}$$
$$a(0.5) = -0.1pi^2cos(pi imes 0.5) = -0.1pi^2cos(pi/2) = -0.1pi^2 imes 0 = 0 ext{ m/s}^2$$
Verification: At the equilibrium position ($x=0$), velocity is maximum (or minimum in magnitude here, as it's moving towards negative extreme) and acceleration is zero. Our results match this perfectly.
d) Kinetic, potential, and total energy at $t = 0.5$ s:
We need the force constant 'k'.
We know $omega^2 = k/m Rightarrow k = momega^2$.
$k = 0.1 ext{ kg} imes (pi ext{ rad/s})^2 = 0.1pi^2 ext{ N/m}$.
At $t = 0.5$ s, we found $x = 0$ m and $v = -0.1pi$ m/s.
Kinetic Energy (KE):
$$KE = frac{1}{2}mv^2 = frac{1}{2} imes 0.1 ext{ kg} imes (-0.1pi ext{ m/s})^2$$
$$KE = frac{1}{2} imes 0.1 imes 0.01pi^2 = 0.0005pi^2 ext{ J}$$
$$KE approx 0.0005 imes (3.14)^2 approx 0.0005 imes 9.8596 approx 0.00493 ext{ J}$$
Potential Energy (PE):
$$PE = frac{1}{2}kx^2 = frac{1}{2} imes (0.1pi^2 ext{ N/m}) imes (0 ext{ m})^2 = 0 ext{ J}$$
Total Mechanical Energy (TME):
$$TME = KE + PE = 0.0005pi^2 ext{ J} + 0 ext{ J} = 0.0005pi^2 ext{ J}$$
Alternatively, we can calculate TME using $TME = frac{1}{2}kA^2$:
$$TME = frac{1}{2} imes (0.1pi^2 ext{ N/m}) imes (0.1 ext{ m})^2$$
$$TME = frac{1}{2} imes 0.1pi^2 imes 0.01 = 0.0005pi^2 ext{ J}$$
The values match, demonstrating the conservation of total mechanical energy in SHM!
---
This detailed exploration of SHM equations and energy dynamics should equip you with a strong foundation for tackling complex problems in JEE. Remember, practice is key to mastering these concepts. Keep working on problems, and you'll find SHM to be a very predictable and elegant topic!