Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Oscillations & Simple Harmonic Motion for CBSE Class 11 Physics.
Key Concepts
Simple Harmonic Motion (SHM)
The mathematics of repetitive back-and-forth movement
1
Periodic vs. Oscillatory Motion
Definition
Periodic motion repeats itself after equal intervals of time (e.g., Earth around the Sun). Oscillatory motion is periodic motion that moves to and fro about a mean position (e.g., a pendulum).
$$\text{All oscillatory motion is periodic, but not all periodic motion is oscillatory.}$$
2
Condition for SHM
Formula
A specific type of oscillatory motion where the restoring force ($F$) is directly proportional to the displacement ($x$) from the mean position, and always directed towards it.
$$F = -kx \quad \text{or} \quad a = -\omega^2 x$$
3
Displacement in SHM
Formula
The position of the particle at any time $t$. It is represented by a sine or cosine function. $A$ is the amplitude, $\omega$ is angular frequency, and $\phi$ is the initial phase.
$$x(t) = A \cos(\omega t + \phi) \quad \text{or} \quad x(t) = A \sin(\omega t + \phi)$$
4
Velocity and Acceleration in SHM
Formula
Found by differentiating displacement. Velocity is maximum at the mean position ($x=0$) and zero at the extremes. Acceleration is zero at the mean and maximum at the extremes.
$$v = \pm \omega \sqrt{A^2 – x^2} \quad | \quad v_{\text{max}} = A\omega$$
$$a = -\omega^2 x \quad | \quad a_{\text{max}} = A\omega^2$$
5
Energy in SHM
Formula
The system constantly trades Kinetic Energy ($K$) for Potential Energy ($U$). In an undamped system, Total Mechanical Energy ($E$) remains perfectly constant and is proportional to the square of the amplitude.
$$K = \frac{1}{2} m \omega^2 (A^2 – x^2)$$
$$U = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} k x^2$$
$$E = K + U = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} k A^2$$
6
Time Period of a Spring-Mass System
Formula
The time taken to complete one full oscillation. For a mass $m$ attached to a spring of constant $k$, the period depends only on mass and stiffness, NOT on amplitude or gravity.
$$T = 2\pi \sqrt{\frac{m}{k}}$$
7
Time Period of a Simple Pendulum
Formula
For a small angular displacement ($\theta < 10^\circ$), a pendulum executes SHM. The period depends only on the length ($L$) and gravity ($g$), NOT on the mass of the bob.
$$T = 2\pi \sqrt{\frac{L}{g}}$$
8
Damped Oscillations
Concept
In the real world, friction and air drag cause the amplitude of oscillation to decrease exponentially over time. Mechanical energy is dissipated as heat.
$$x(t) = A e^{-bt/2m} \cos(\omega’ t + \phi)$$
9
Forced Oscillations & Resonance
Definition
When an external periodic force drives a system. If the driving frequency ($\omega_d$) exactly matches the system’s natural frequency ($\omega_0$), the amplitude of oscillation becomes dangerously large. This is Resonance.
$$\text{Resonance occurs when } \omega_d = \omega_0$$
Concept Deep Dive
01
The Reference Circle Method
SHM is just Circular Motion in disguiseImagine a particle moving in a perfect circle at a constant speed. Now, shine a light on it from the side and watch its shadow on the wall.
As the particle goes round and round, the shadow simply moves strictly up and down in a straight line. That shadow is executing perfect Simple Harmonic Motion. This is why we use $\sin$ and $\cos$, and why we measure frequency in radians per second ($\omega$), even though the spring isn’t moving in a circle. The mathematics of SHM are identical to the 1D projection of uniform circular motion!
As the particle goes round and round, the shadow simply moves strictly up and down in a straight line. That shadow is executing perfect Simple Harmonic Motion. This is why we use $\sin$ and $\cos$, and why we measure frequency in radians per second ($\omega$), even though the spring isn’t moving in a circle. The mathematics of SHM are identical to the 1D projection of uniform circular motion!
02
What is Initial Phase ($\phi$)?
Setting the clockStudents often get confused by $x = A \sin(\omega t + \phi)$. What is $\phi$?
It simply tells you where the particle was when you started your stopwatch ($t=0$).
– If you start timing exactly as it passes the mean position moving positively, $\phi = 0$ (so $x = A \sin \omega t$).
– If you pull a spring to its maximum stretch and then start timing, the particle is starting at the extreme. This corresponds to $\phi = \pi/2$ (so $x = A \sin(\omega t + \pi/2) = A \cos \omega t$).
It simply tells you where the particle was when you started your stopwatch ($t=0$).
– If you start timing exactly as it passes the mean position moving positively, $\phi = 0$ (so $x = A \sin \omega t$).
– If you pull a spring to its maximum stretch and then start timing, the particle is starting at the extreme. This corresponds to $\phi = \pi/2$ (so $x = A \sin(\omega t + \pi/2) = A \cos \omega t$).
Compare & Contrast
✗ Springs in Series
- Connected end-to-end.
- Tension/Force is the same in all springs.
- Total extension is the sum of individual extensions.
- Effective spring constant decreases.
- Formula: $\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}$
✓ Springs in Parallel
- Connected side-by-side to the same mass.
- Extension ($x$) is the same for all springs.
- Total force is the sum of individual forces.
- Effective spring constant increases (stiffer).
- Formula: $k_{\text{eq}} = k_1 + k_2$
Remember
If a block is placed *between* two springs (attached to opposite walls), they behave as if they are in Parallel, not series! When the block moves right, the right spring pushes back, and the left spring pulls back—both forces act in the same restoring direction, making it stiffer ($k_1 + k_2$).
Common Mistakes to Avoid
Mistake 1
Assuming Time Period depends on Amplitude: If you pull a pendulum back $2^\circ$ or $5^\circ$, it will take the exact same time to swing back and forth. The larger swing travels faster, perfectly compensating for the extra distance. $T$ is independent of $A$.
Mistake 2
Misjudging Velocity at the Mean Position: Students sometimes think acceleration and velocity peak together. In SHM, they are out of phase. When a pendulum swings through the very bottom (mean position), it is moving at its absolute maximum speed ($v = A\omega$), but its acceleration is completely zero!
Mistake 3
Using the Pendulum formula in Free Fall: The formula $T = 2\pi\sqrt{L/g}$ relies on gravity providing a restoring force. If the pendulum is in a satellite or a freely falling elevator, the effective gravity is zero ($g_{\text{eff}} = 0$). Therefore, the time period becomes infinity—it will not oscillate at all!
Exam Tips
Tip 1
The Double Frequency Rule: If a particle performs SHM with a frequency $\nu$, its Kinetic Energy and Potential Energy oscillate with a frequency of $2\nu$. (Because energy depends on $v^2$ and $x^2$, it peaks twice during a single full mechanical oscillation).
Tip 2
Where is KE = PE? This is a classic 1-mark question. Set $\frac{1}{2}m\omega^2(A^2 – x^2) = \frac{1}{2}m\omega^2 x^2$. Solving this gives $A^2 – x^2 = x^2 \implies 2x^2 = A^2 \implies x = \pm \frac{A}{\sqrt{2}} \approx \pm 0.707 A$. The energies are equal at roughly 71% of the maximum amplitude.
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Oscillations · CBSE Exam1
A spring of force constant $k$ is cut into two equal halves. What will be the force constant of each half? [1 mark]
Answer
$2k$
📝
Explanation
The spring constant $k$ is inversely proportional to the unstretched length of the spring ($k \propto 1/L$). If the spring is cut in half, its length becomes $L/2$. Therefore, the stiffness of the spring doubles, making the new force constant $2k$.
2
A particle executes SHM of amplitude $A$. At what distance from the mean position is its kinetic energy equal to its potential energy? [2 marks]
Answer
$x = \pm A / \sqrt{2}$
📝
Explanation
Kinetic Energy $K = \frac{1}{2} m \omega^2 (A^2 – x^2)$
Potential Energy $U = \frac{1}{2} m \omega^2 x^2$
Given $K = U$:
$\frac{1}{2} m \omega^2 (A^2 – x^2) = \frac{1}{2} m \omega^2 x^2$
Cancelling the common terms: $A^2 – x^2 = x^2$
$A^2 = 2x^2 \implies x^2 = A^2 / 2 \implies x = \pm \frac{A}{\sqrt{2}}$.
3
The equation of a particle executing SHM is given by $x = 5 \sin(\pi t + \pi/3)$ in SI units. Calculate the (a) amplitude, (b) time period, and (c) maximum velocity of the particle. [3 marks]
Answer
(a) $5 \text{ m}$, (b) $2 \text{ s}$, (c) $5\pi \text{ m/s}$
📝
Explanation
Comparing the given equation with the standard equation $x = A \sin(\omega t + \phi)$:
(a) Amplitude ($A$): Direct comparison gives $A = 5 \text{ m}$.
(b) Time Period ($T$): We see $\omega = \pi$. Since $\omega = 2\pi / T$, we get $\pi = 2\pi / T \implies T = 2 \text{ seconds}$.
(c) Maximum Velocity: The formula is $v_{\text{max}} = A\omega$.
$v_{\text{max}} = 5 \times \pi = 5\pi \text{ m/s}$ (approx $15.7 \text{ m/s}$).
Concept Map
Oscillations connects to →
Periodic Physics
Waves (Propagation of SHM)
Circular Motion (Reference Circle)
Alternating Current (Class 12 analogy)
Sound (Acoustic resonance)
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