Spring-Mass System — Revision Notes

⚡ 30-Second Revision

Hooke's Law: — $F = -kx$
Angular Frequency: — $\omega = \sqrt{\frac{k}{m}}$
Time Period: — $T = 2\pi\sqrt{\frac{m}{k}}$
Frequency: — $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$
Total Energy: — $E = \frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
Springs in Series: — $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + ...$
Springs in Parallel: — $k_{eq} = k_1 + k_2 + ...$
Spring Constant & Length: — $k \propto 1/L$ (if cut, $k_{new} = k_{original} \times (L_{original}/L_{new})$)
Vertical System: — $T$ is independent of $g$. Equilibrium position shifts.

2-Minute Revision

The spring-mass system is the classic example of Simple Harmonic Motion (SHM). Its behavior is governed by Hooke's Law, $F = -kx$, where $k$ is the spring constant and $x$ is the displacement from equilibrium.

This restoring force leads to oscillations with an angular frequency $\omega = \sqrt{k/m}$, a time period $T = 2\pi\sqrt{m/k}$, and frequency $f = \frac{1}{2\pi}\sqrt{k/m}$. Remember that the time period depends only on the mass and spring constant, not on the amplitude or gravity (for vertical systems, gravity only shifts the equilibrium point).

Energy is conserved in an ideal system, continuously converting between kinetic energy ($K = \frac{1}{2}mv^2$) and potential energy ($U = \frac{1}{2}kx^2$). The total energy is $E = \frac{1}{2}kA^2$, where $A$ is the amplitude.

For combinations, springs in series have $1/k_{eq} = 1/k_1 + 1/k_2$, while in parallel, $k_{eq} = k_1 + k_2$. If a spring is cut, its constant changes inversely with length. These are the most testable concepts for NEET.

5-Minute Revision

The spring-mass system is fundamental to understanding oscillations. At its core is Hooke's Law, $F = -kx$, where $F$ is the restoring force, $k$ is the spring constant (stiffness), and $x$ is the displacement from equilibrium.

The negative sign indicates the force opposes displacement. This leads to the SHM differential equation $m\frac{d^2x}{dt^2} = -kx$, from which we derive the angular frequency $\omega = \sqrt{k/m}$. Consequently, the time period $T = 2\pi/\omega = 2\pi\sqrt{m/k}$ and frequency $f = 1/T = \frac{1}{2\pi}\sqrt{k/m}$.

Key Points & Formulas:

1
Time Period: — $T = 2\pi\sqrt{m/k}$. This is independent of amplitude and, for vertical systems, independent of $g$.
2
Energy Conservation: — Total mechanical energy $E = K + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$ is conserved. At maximum displacement ($x=A$, $v=0$), $E = \frac{1}{2}kA^2$. At equilibrium ($x=0$, $v=v_{max}$), $E = \frac{1}{2}mv_{max}^2$.

* Example: If $E=100,\text{J}$ and $U=25,\text{J}$, then $K = E-U = 75,\text{J}$.

1
Spring Combinations:

* Series: $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$. The extension adds up, force is same. * Parallel: $k_{eq} = k_1 + k_2$. The forces add up, extension is same. * Example: Two springs $k_1=100,\text{N/m}$, $k_2=200,\text{N/m}$ in parallel: $k_{eq}=300,\text{N/m}$. In series: $k_{eq} = (100 \times 200)/(100+200) = 20000/300 = 200/3,\text{N/m}$.

1
Cutting a Spring: — If a spring of constant $k$ and length $L$ is cut into $n$ equal parts, each part has a constant $nk$. Generally, $k \propto 1/L$.

* Example: A spring $k=100,\text{N/m}$ cut into 4 equal parts. Each part has $k'=400,\text{N/m}$.

Focus on these variations and energy transformations. Practice numerical problems to solidify your understanding.

Prelims Revision Notes

The spring-mass system is a key topic for NEET, primarily testing your understanding of Simple Harmonic Motion (SHM) and its associated formulas.

1. Basic Definitions & Formulas:

* Restoring Force: $F = -kx$. $k$ is spring constant (N/m), $x$ is displacement (m). * Angular Frequency: $\omega = \sqrt{k/m}$ (rad/s). * Time Period: $T = 2\pi\sqrt{m/k}$ (s). This is the time for one complete oscillation. * Frequency: $f = 1/T = \frac{1}{2\pi}\sqrt{k/m}$ (Hz).

2. Energy in SHM:

* Kinetic Energy: $K = \frac{1}{2}mv^2$. * Potential Energy (Elastic): $U = \frac{1}{2}kx^2$. * Total Mechanical Energy: $E = K + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$. This is constant in an ideal system. * At extreme positions ($x=\pm A$, $v=0$): $K=0$, $U = \frac{1}{2}kA^2 = E$. * At equilibrium position ($x=0$, $v=v_{max}$): $U=0$, $K = \frac{1}{2}mv_{max}^2 = E$.

3. Spring Combinations:

* Series: $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + ...$. (Think of resistors in parallel). * Parallel: $k_{eq} = k_1 + k_2 + ...$. (Think of resistors in series).

4. Effect of Cutting a Spring:

* Spring constant $k$ is inversely proportional to its length $L$ ($kL = \text{constant}$). If a spring is cut into $n$ equal parts, each part has a spring constant $nk$.

5. Vertical Spring-Mass System:

* The time period $T = 2\pi\sqrt{m/k}$ remains unchanged. It is independent of $g$. * The equilibrium position shifts downwards by $\Delta L = mg/k$.

6. System in Accelerating Lift:

* Time period $T$ is still independent of lift's acceleration. Only the equilibrium position shifts due to change in effective weight.

7. Common Traps:

* Confusing series/parallel formulas. * Incorrectly calculating $k$ after cutting a spring. * Errors in energy calculations at specific displacements. * Forgetting to take the square root in $T$ or $\omega$ calculations.

Vyyuha Quick Recall

To remember the time period formula: 'Two Pi, Root M over K' (sounds like 'Two Pie, Root M over K'). This helps recall $T = 2\pi\sqrt{m/k}$. For spring combinations: 'Series is Sum of Reciprocals, Parallel is Plus' (like resistors, but opposite for $k$). So, for series $1/k_{eq} = 1/k_1 + 1/k_2$, and for parallel $k_{eq} = k_1 + k_2$.