AP Physics C Mechanics- 2.8 Spring Forces- Study Notes- New Syllabus
AP Physics C Mechanics- 2.8 Spring Forces – Study Notes
AP Physics C Mechanics- 2.8 Spring Forces – Study Notes – per latest Syllabus.
Key Concepts:
- Spring Forces and Hooke’s Law
- Equivalent Spring Constant for Springs in Series
- Equivalent Spring Constant for Springs in Parallel
Ideal and Nonideal Springs
An ideal spring is a spring that has negligible mass and obeys Hooke’s Law perfectly — meaning the restoring force it exerts is directly proportional to its change in length from its relaxed (unstretched) position.
\( \mathrm{F_s \propto \Delta x} \)
- Relaxed length: The natural length of the spring when no external force is applied.
- Extension or compression: The change in the spring’s length from this relaxed length.
Nonideal springs deviate from this behavior. They may:
- Have non-negligible mass (so the spring’s own weight affects its behavior).
- Exert a nonlinear force (e.g., rubber bands or springs with coil deformation at large stretches).
Key Idea: Ideal springs obey a perfect linear relationship between force and extension, while nonideal springs do not.
Hooke’s Law
The restoring force exerted by an ideal spring on an attached object is described by Hooke’s Law:
\( \mathrm{\vec{F}_s = -k \, \Delta \vec{x}} \)
- \( \mathrm{\vec{F}_s} \): spring (restoring) force (N)
- \( \mathrm{k} \): spring constant (N/m), measures stiffness
- \( \mathrm{\Delta \vec{x}} \): displacement from the equilibrium or relaxed length (m)
Explanation:
- The negative sign indicates that the spring’s force is directed opposite to the displacement — it always acts to restore the spring to equilibrium.
- If the spring is stretched (\( \mathrm{\Delta x > 0} \)), the force acts toward the spring.
- If the spring is compressed (\( \mathrm{\Delta x < 0} \)), the force acts away from the compression direction.
Key Idea: The stiffer the spring (larger \( \mathrm{k} \)), the greater the restoring force for the same displacement.
Direction of Spring Force
The force exerted by a spring on an object is always directed toward the equilibrium position of the system.
Explanation:
- When stretched, the spring pulls the object back toward equilibrium.
- When compressed, it pushes the object outward toward equilibrium.
- This restoring tendency is what allows springs to produce oscillatory motion in systems such as mass–spring oscillators.
Key Idea: The spring’s force always acts to oppose the displacement, maintaining equilibrium of the object–spring system.
Example
A horizontal spring with a spring constant of \( \mathrm{k = 250\,N/m} \) is attached to a wall. A block is connected to the other end of the spring and pulled horizontally so that the spring is stretched by \( \mathrm{0.12\,m.} \)
Find the magnitude and direction of the restoring force exerted by the spring on the block.
▶️ Answer / Explanation
Step 1: Write Hooke’s Law:
\( \mathrm{F_s = -k\,\Delta x} \)
Step 2: Substitute given values:
\( \mathrm{F_s = -(250)(0.12)} \)
\( \mathrm{F_s = -30\,N.} \)
Step 3: Interpret the negative sign:
The negative sign indicates that the spring’s restoring force acts opposite the direction of displacement — it pulls the block back toward equilibrium.
Physical Meaning: A stronger spring (larger \( \mathrm{k} \)) or a greater stretch (\( \mathrm{\Delta x} \)) would produce a larger restoring force. This relationship defines the behavior of an ideal spring under Hooke’s Law.
Equivalent Spring Constant for Springs in Series
When multiple springs act together on the same object, the system can often be replaced by a single equivalent spring that produces the same overall restoring force for a given displacement. This equivalent spring has an effective spring constant \( \mathrm{k_{eq}} \), which depends on how the springs are connected — in series or in parallel.
Springs in Series
When springs are connected in series (end-to-end), they share the same force, but the total extension of the system is the sum of the individual extensions of each spring.
Derived Relationship:
\( \mathrm{\dfrac{1}{k_{eq}} = \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dfrac{1}{k_3} + \dots} \)
Explanation:
- The same force \( \mathrm{F} \) acts on each spring in the series connection.
- Each spring stretches by a different amount depending on its stiffness \( \mathrm{k_i} \): \( \mathrm{x_i = \dfrac{F}{k_i}} \).
- The total stretch \( \mathrm{x_{total}} \) is the sum of individual stretches:
\( \mathrm{x_{total} = \dfrac{F}{k_1} + \dfrac{F}{k_2} + \dots = F \left(\dfrac{1}{k_1} + \dfrac{1}{k_2} + \dots \right)} \)
- Using \( \mathrm{F = k_{eq} x_{total}} \), the inverse relationship for \( \mathrm{k_{eq}} \) follows.
In a series arrangement, the equivalent spring constant is always less than the smallest individual spring constant. The system becomes “softer” because the springs collectively stretch more under the same force.
Example
Two springs are connected in series and attached to a block on a horizontal frictionless surface. The first spring has a spring constant \( \mathrm{k_1 = 300\,N/m} \), and the second has \( \mathrm{k_2 = 150\,N/m.} \) Determine the equivalent spring constant \( \mathrm{k_{eq}} \) of the combination and the total extension if a \( \mathrm{6.0\,N} \) force is applied.
▶️ Answer / Explanation
Step 1: Use the formula for springs in series:
\( \mathrm{\dfrac{1}{k_{eq}} = \dfrac{1}{k_1} + \dfrac{1}{k_2}} \)
\( \mathrm{\dfrac{1}{k_{eq}} = \dfrac{1}{300} + \dfrac{1}{150} = \dfrac{1 + 2}{300} = \dfrac{3}{300}} \)
\( \mathrm{k_{eq} = 100\,N/m.} \)
Step 2: Calculate total extension under \( \mathrm{F = 6.0\,N.} \)
\( \mathrm{x_{total} = \dfrac{F}{k_{eq}} = \dfrac{6.0}{100} = 0.06\,m = 6.0\,cm.} \)
Step 3: Optional – Individual spring extensions:
- \( \mathrm{x_1 = \dfrac{F}{k_1} = \dfrac{6}{300} = 0.020\,m = 2.0\,cm.} \)
- \( \mathrm{x_2 = \dfrac{F}{k_2} = \dfrac{6}{150} = 0.040\,m = 4.0\,cm.} \)
- \( \mathrm{x_{total} = x_1 + x_2 = 2.0 + 4.0 = 6.0\,cm.} \)
Result:
- Equivalent spring constant: \( \mathrm{k_{eq} = 100\,N/m.} \)
- Total extension: \( \mathrm{0.06\,m = 6.0\,cm.} \)
- The system behaves like a single spring that is softer than either spring alone.
Equivalent Spring Constant for Springs in Parallel
When multiple springs are connected in parallel (side by side), each spring experiences the same displacement, but the total restoring force is the sum of the forces produced by each spring.
Derived Relationship:
\( \mathrm{k_{eq} = k_1 + k_2 + k_3 + \dots} \)
Explanation:
- Each spring supports part of the total force.
- Since the displacements are identical, the forces add directly to form a larger restoring force for a given extension.
- The system behaves as a stiffer spring — the more springs in parallel, the greater the equivalent spring constant.
Connecting springs in parallel makes the system stiffer, while connecting them in series makes it softer.
Example
Two springs, one with \( \mathrm{k_1 = 300\,N/m} \) and the other with \( \mathrm{k_2 = 200\,N/m,} \) are connected first in series and then in parallel. Determine how the stiffness of the system changes in each case and explain why.
▶️ Answer / Explanation
Step 1: For series connection, the equivalent spring constant is found from:
\( \mathrm{\dfrac{1}{k_{eq}} = \dfrac{1}{k_1} + \dfrac{1}{k_2}} \)
\( \mathrm{\dfrac{1}{k_{eq}} = \dfrac{1}{300} + \dfrac{1}{200} = \dfrac{5}{600}} \)
\( \mathrm{k_{eq} = 120\,N/m.} \)
Step 2: For parallel connection, the equivalent spring constant is the sum of the individual constants:
\( \mathrm{k_{eq} = k_1 + k_2 = 300 + 200 = 500\,N/m.} \)
Step 3: Interpretation:
- In series, the springs stretch together, sharing the same force, and the total extension increases — the system behaves like a softer spring.
- In parallel, each spring supports part of the load, so the system resists deformation more — it behaves like a stiffer spring.
Result: The series arrangement yields a weaker equivalent stiffness (\( \mathrm{120\,N/m} \)), while the parallel arrangement is stronger (\( \mathrm{500\,N/m} \)). This shows how spring configuration dramatically affects the system’s overall mechanical response.