Oscillations of Spring — Revision Notes

⚡ 30-Second Revision

Hooke's Law: — $F = -kx$
Angular Frequency: — $\omega = \sqrt{k/m}$
Period: — $T = 2pisqrt{m/k}$
Frequency: — $f = \frac{1}{2pi}\sqrt{k/m}$
Displacement: — $x(t) = Acos(\omega t + \phi)$
Velocity: — $v(t) = -Aomegasin(\omega t + \phi)$
Maximum Velocity: — $v_{max} = Aomega$
Acceleration: — $a(t) = -Aomega^2\cos(\omega t + \phi) = -\omega^2 x$
Maximum Acceleration: — $a_{max} = Aomega^2$
Potential Energy: — $U = \frac{1}{2}kx^2$
Kinetic Energy: — $K = \frac{1}{2}mv^2$
Total Mechanical Energy: — $E = \frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$
Springs in Series: — $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + \dots$
Springs in Parallel: — $k_{eq} = k_1 + k_2 + \dots$
Period is independent of amplitude and gravity (for vertical spring, only equilibrium shifts).

2-Minute Revision

The oscillations of a spring-mass system are a prime example of Simple Harmonic Motion (SHM). The core principle is Hooke's Law, $F = -kx$, where the restoring force is proportional to displacement $x$ and acts oppositely.

This leads to the defining SHM equation $a = -\omega^2 x$, with angular frequency $\omega = \sqrt{k/m}$. The period of oscillation, $T = 2pisqrt{m/k}$, is crucial and depends only on mass ($m$) and spring constant ($k$), not on amplitude or gravity (for vertical springs, gravity just shifts the equilibrium point).

Frequency $f = 1/T$. Energy is conserved, continuously converting between kinetic ($K = \frac{1}{2}mv^2$) and potential ($U = \frac{1}{2}kx^2$) forms. Total energy is constant, $E = \frac{1}{2}kA^2$. Maximum velocity ($Aomega$) occurs at equilibrium, and maximum acceleration ($Aomega^2$) at extreme positions.

Remember how spring constants combine: series ($\frac{1}{k_{eq}} = \sum \frac{1}{k_i}$) makes the system 'softer', parallel ($k_{eq} = \sum k_i$) makes it 'stiffer'. Be wary of common traps like confusing spring and pendulum dependencies or miscalculating energy at specific points.

5-Minute Revision

Let's quickly review the essentials of spring oscillations for NEET. At its heart, a spring-mass system undergoes Simple Harmonic Motion (SHM) because the restoring force, $F = -kx$ (Hooke's Law), is directly proportional to the displacement $x$ from equilibrium and acts in the opposite direction.

This leads to an acceleration $a = -(k/m)x$, which is the hallmark of SHM. From this, we derive the angular frequency $\omega = \sqrt{k/m}$, the period $T = 2pisqrt{m/k}$, and the frequency $f = \frac{1}{2pi}\sqrt{k/m}$.

Key takeaways for $T$ and $f$: They depend on mass and spring constant, but not on the amplitude of oscillation. For vertical springs, gravity shifts the equilibrium position but does not change the period. If you cut a spring, its constant changes inversely with length (e.g., half length means double $k$).

Energy conservation is vital. The total mechanical energy $E = K + U$ is constant. Kinetic energy $K = \frac{1}{2}mv^2$ is maximum at equilibrium ($x=0$), where $v = v_{max} = Aomega$. Potential energy $U = \frac{1}{2}kx^2$ is maximum at the extreme positions ($x=\pm A$), where $v=0$. The total energy can be expressed as $E = \frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$. This allows you to find velocity at any displacement: $v = omegasqrt{A^2 - x^2}$.

Example: A $0.1\,\text{kg}$ mass oscillates on a spring with $k = 40\,\text{N/m}$ and amplitude $0.05\,\text{m}$.

1
Period: — $T = 2pisqrt{0.1/40} = 2pisqrt{1/400} = 2pi/20 = pi/10\,\text{s}$.
2
Total Energy: — $E = \frac{1}{2}kA^2 = \frac{1}{2} \times 40 \times (0.05)^2 = 20 \times 0.0025 = 0.05\,\text{J}$.
3
Maximum Speed: — $E = \frac{1}{2}mv_{max}^2 \implies 0.05 = \frac{1}{2} \times 0.1 \times v_{max}^2 \implies v_{max}^2 = 1 \implies v_{max} = 1\,\text{m/s}$.

Remember spring combinations: for series, $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$; for parallel, $k_{eq} = k_1 + k_2$. Practice these formulas and conceptual nuances to ace NEET questions.

Prelims Revision Notes

Oscillations of Spring: NEET Revision Notes

1. Simple Harmonic Motion (SHM) Basics:

Definition: — A type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction.
Hooke's Law: — $F = -kx$, where $k$ is the spring constant (N/m) and $x$ is displacement.
SHM Equation: — $a = -\omega^2 x$, where $\omega$ is angular frequency.

2. Key Parameters & Formulas:

Angular Frequency ($\omega$): — $\omega = \sqrt{k/m}$ (rad/s)
Period (T): — Time for one complete oscillation. $T = 2pi/\omega = 2pisqrt{m/k}$ (s)

* Crucial: $T$ is independent of amplitude for ideal SHM. * Crucial: For vertical springs, $T$ is independent of gravity ($g$ only shifts equilibrium).

Frequency (f): — Number of oscillations per second. $f = 1/T = omega/(2\pi) = \frac{1}{2pi}\sqrt{k/m}$ (Hz)

3. Displacement, Velocity, and Acceleration:

Displacement: — $x(t) = Acos(\omega t + \phi)$ (or $Asin(\omega t + \phi)$)
Velocity: — $v(t) = -Aomegasin(\omega t + \phi)$. Max velocity $v_{max} = Aomega$ (at $x=0$).
Acceleration: — $a(t) = -Aomega^2\cos(\omega t + \phi) = -\omega^2 x$. Max acceleration $a_{max} = Aomega^2$ (at $x=\pm A$).

4. Energy in SHM (Conservation of Mechanical Energy):

Elastic Potential Energy (U): — $U = \frac{1}{2}kx^2$. Maximum at $x=\pm A$, minimum (0) at $x=0$.
Kinetic Energy (K): — $K = \frac{1}{2}mv^2$. Maximum at $x=0$, minimum (0) at $x=\pm A$.
Total Mechanical Energy (E): — $E = K + U = \frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$ (constant).
Velocity at displacement x: — $v = \pm omegasqrt{A^2 - x^2}$.
When $K=U$: — Occurs at $x = \pm A/\sqrt{2}$.

5. Combinations of Springs:

Series Connection: — Springs connected end-to-end. $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + \dots$. $k_{eq}$ is smaller than individual $k$'s (softer system, longer period).
Parallel Connection: — Springs connected side-by-side, sharing displacement. $k_{eq} = k_1 + k_2 + \dots$. $k_{eq}$ is larger than individual $k$'s (stiffer system, shorter period).

6. Spring Constant and Length:

For a spring, $k \propto 1/L$. If a spring is cut into $n$ equal parts, each part has spring constant $nk$.

7. Mass of Spring:

If spring mass $m_s$ is considered, effective mass $m_{eff} = m_{block} + m_s/3$. Then $T = 2pisqrt{m_{eff}/k}$.

8. Common Traps:

Confusing spring-mass period dependencies with simple pendulum dependencies (e.g., effect of gravity, mass).
Incorrectly applying series/parallel formulas.
Assuming linear relationships for energy or velocity with displacement.

Vyyuha Quick Recall

To remember the period formula for a spring-mass system, think: 'Two Pi, My King!'

Two Pi — $\rightarrow 2\pi$
My — $\rightarrow m$ (mass)
King — $\rightarrow k$ (spring constant)

So, $T = 2pisqrt{\frac{\text{My}}{\text{King}}} \implies T = 2pisqrt{\frac{m}{k}}$