Forces - Physics Fun

Springs

A spring exerts a force on whatever is attached to it whenever it is stretched or compressed away from its natural length. The further you displace it, the harder it pushes back. Everything else on this page follows from that single idea.

Hooke's Law

The force a spring exerts is proportional to its displacement from equilibrium. This is Hooke's Law:

\( F_s = -kx \)

Where:

\( F_s \) — spring force (N). Always directed back toward equilibrium.
\( k \) — spring constant (N/m). Larger \( k \) = stiffer spring.
\( x \) — displacement from equilibrium (m). Positive = stretched, negative = compressed.

The negative sign means the spring force always opposes the displacement — pull it right and it pulls back left. This is called a restoring force. On a Force vs. Displacement graph, the slope of the line equals \( k \), and the area under the curve equals the work done on the spring.

Spring Potential Energy

A displaced spring stores energy. Since the force varies linearly with displacement, the energy stored goes as \( x^2 \) — doubling the stretch stores four times the energy.

\( U_s = \frac{1}{2}kx^2 \)

Where:

\( U_s \) — elastic potential energy (J). Always ≥ 0.
\( k \) — spring constant (N/m)
\( x \) — displacement from equilibrium (m)

💡 Energy Scales as x², Not x Spring PE goes as x² — doubling the stretch stores four times the energy, not twice. This is because the spring force increases linearly with stretch, so more work is done per meter the further you pull. It shows up on Force vs. Displacement graphs as the area under a triangle.

Oscillation: Period and Frequency

When a mass is attached to a spring and released, it oscillates back and forth. Two key terms:

Period (T) — time for one complete oscillation (seconds)
Frequency (f) — oscillations per second (Hz)

\( T = \frac{1}{f} \qquad f = \frac{1}{T} \)

The period of a spring-mass system is:

\( T = 2\pi\sqrt{\frac{m}{k}} \)

Where:

\( T \) — period (s)
\( m \) — attached mass (kg)
\( k \) — spring constant (N/m)

Three things AP exams test heavily about this equation:

More mass → longer period. More inertia means slower oscillation.
Stiffer spring → shorter period. A stronger restoring force speeds things up.
Amplitude does NOT affect the period. Pull it farther and it travels faster by exactly the right amount to take the same time. This is non-obvious and frequently tested.

⚠ Common Mistake Amplitude does not affect the period of a spring-mass system. A larger amplitude means more distance to cover, but the mass also moves faster — these effects cancel exactly. Period depends only on m and k. This surprises most students and is heavily tested.

Energy in a Spring System

In a frictionless spring system, total mechanical energy is conserved — it continuously trades between kinetic energy and spring PE:

\( E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant} \)

The two key positions:

At maximum displacement \( x = A \): the mass stops momentarily — all energy is stored as spring PE: \( E = \frac{1}{2}kA^2 \)
At equilibrium \( x = 0 \): all PE has converted to KE — the mass moves fastest here: \( E = \frac{1}{2}mv_{\text{max}}^2 \)

Setting those equal gives the maximum speed:

\( v_{\text{max}} = A\sqrt{\frac{k}{m}} \)

At any other position \( x \), use conservation directly:

\( \frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \)

Example

A 0.5 kg block is attached to a spring (\( k = 200\ \text{N/m} \)) and pulled 0.10 m from equilibrium, then released from rest.

Period: \( T = 2\pi\sqrt{0.5/200} = 2\pi(0.05) \approx 0.314\ \text{s} \)

Maximum speed: \( v_{\text{max}} = 0.10\sqrt{200/0.5} = 0.10(20) = 2.0\ \text{m/s} \)

Speed at \( x = 0.06\ \text{m} \):
\( 1.0\ \text{J} = \frac{1}{2}(0.5)v^2 + \frac{1}{2}(200)(0.06)^2 = 0.25v^2 + 0.36 \)
\( v = \sqrt{0.64/0.25} = 1.6\ \text{m/s} \)

Practice Problems

A spring with \( k = 400\ \text{N/m} \) is compressed 0.05 m. How much energy is stored in the spring?
1. 0.25 J
2. 0.50 J
3. 1.0 J
4. 10 J

A 2 kg block oscillates on a spring. If the spring constant is doubled and the mass is quadrupled, how does the period change?
1. Period doubles
2. Period is halved
3. Period stays the same
4. Period increases by a factor of \( \sqrt{2} \)

A mass on a spring is pulled to amplitude A and released. At what position does the mass have the greatest kinetic energy?
1. At \( x = A \)
2. At \( x = A/2 \)
3. At \( x = 0 \) (equilibrium)
4. KE is constant throughout the motion

A 3 kg block hangs vertically from a spring and stretches it 0.12 m from its natural length when the system is at rest. What is the spring constant?
1. 25 N/m
2. 245 N/m
3. 36 N/m
4. 3.6 N/m

A 2 kg block is held against a spring (\( k = 500\ \text{N/m} \)) compressed 0.08 m on a frictionless horizontal surface and released. What is the speed of the block when it leaves the spring?
1. 1.0 m/s
2. 1.26 m/s
3. 2.0 m/s
4. 4.0 m/s

Combination — Ramp + Spring

A 4 kg block is placed on a frictionless ramp inclined at 30° and pushed against a spring at the bottom of the ramp, compressing it by 0.20 m. When released, the block just barely reaches a point 0.20 m up the ramp (measured along the ramp) before stopping. What is the spring constant \( k \)?
(\( g = 10\ \text{m/s}^2 \))
1. 50 N/m
2. 100 N/m
3. 200 N/m
4. 400 N/m

Free Response

A 0.8 kg block is attached to a horizontal spring (\( k = 320\ \text{N/m} \)) on a frictionless surface and displaced 0.15 m from equilibrium, then released from rest.
1. Calculate the period of the oscillation.
2. Calculate the total mechanical energy of the system.
3. Calculate the maximum speed of the block.
4. Calculate the speed of the block when it is 0.09 m from equilibrium.
5. Without calculation: if the amplitude is doubled to 0.30 m, how does the period change? How does the maximum speed change? Explain.

Answer Key

Answer Key

Q1 — B (0.50 J)
\( U_s = \frac{1}{2}kx^2 = \frac{1}{2}(400)(0.05)^2 = \frac{1}{2}(400)(0.0025) = 0.50\ \text{J} \)

Q2 — D (period increases by \( \sqrt{2} \))
\( T = 2\pi\sqrt{m/k} \). New mass \( 4m \), new constant \( 2k \):
\( T' = 2\pi\sqrt{\frac{4m}{2k}} = 2\pi\sqrt{\frac{2m}{k}} = \sqrt{2}\cdot 2\pi\sqrt{\frac{m}{k}} = \sqrt{2}\,T \)

Q3 — C (equilibrium, x = 0)
At equilibrium all PE has converted to KE. At the amplitude, KE = 0. KE is maximum where PE is minimum, which is at \( x = 0 \).

Q4 — B (245 N/m)
At equilibrium the spring force equals gravity: \( kx = mg \).
\( k = \frac{mg}{x} = \frac{(3)(9.8)}{0.12} = \frac{29.4}{0.12} \approx \mathbf{245\ \text{N/m}} \)

Q5 — B (1.26 m/s)
All spring PE converts to KE when the block leaves the spring:
\( \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \)
\( \frac{1}{2}(500)(0.08)^2 = \frac{1}{2}(2)v^2 \)
\( 1.6\ \text{J} = v^2 \Rightarrow v = \sqrt{1.6} \approx \mathbf{1.26\ \text{m/s}} \)

Q6 — C (200 N/m)
The spring's energy converts entirely to gravitational PE (frictionless ramp, block stops):
\( \frac{1}{2}kx^2 = mgh \)
Height gained: \( h = d\sin\theta = (0.20)\sin 30° = 0.10\ \text{m} \)
\( \frac{1}{2}k(0.20)^2 = (4)(10)(0.10) \)
\( 0.02k = 4 \Rightarrow k = \mathbf{200\ \text{N/m}} \)

Q7 — Free Response (FRQ) (k = 320 N/m, m = 0.8 kg, A = 0.15 m)

Part A — Period:
\( T = 2\pi\sqrt{\frac{0.8}{320}} = 2\pi\sqrt{0.0025} = 2\pi(0.05) \approx \mathbf{0.314\ \text{s}} \)

Part B — Total energy:
\( E = \frac{1}{2}kA^2 = \frac{1}{2}(320)(0.15)^2 = \frac{1}{2}(320)(0.0225) = \mathbf{3.6\ \text{J}} \)

Part C — Maximum speed:
\( \frac{1}{2}mv_{\text{max}}^2 = 3.6\ \text{J} \Rightarrow v_{\text{max}} = \sqrt{\frac{2(3.6)}{0.8}} = \sqrt{9} = \mathbf{3.0\ \text{m/s}} \)

Part D — Speed at x = 0.09 m:
\( \frac{1}{2}(0.8)v^2 = 3.6 - \frac{1}{2}(320)(0.09)^2 = 3.6 - 1.296 = 2.304\ \text{J} \)
\( v = \sqrt{\frac{2(2.304)}{0.8}} = \sqrt{5.76} = \mathbf{2.4\ \text{m/s}} \)

Part E — Conceptual (A doubled):
Period does not change — period depends only on \( m \) and \( k \), not amplitude.
Maximum speed doubles — \( v_{\text{max}} = A\sqrt{k/m} \), so doubling \( A \) doubles \( v_{\text{max}} \). The total energy quadruples (\( E = \frac{1}{2}kA^2 \)), and since all of it converts to KE at equilibrium, the max speed doubles.

Summary

Springs introduce the idea of a restoring force — a force that always pushes or pulls an object back toward equilibrium. This produces oscillatory motion where energy continuously trades between kinetic and potential forms.

Hooke's Law gives the force: \( F = -kx \). The negative sign means the spring always opposes displacement.
Spring PE grows with the square of displacement: \( U_s = \frac{1}{2}kx^2 \). Doubling displacement stores 4× the energy.
Energy constantly trades between KE and PE. At the amplitude, all energy is PE. At equilibrium, all energy is KE. The total is always constant in a frictionless system.
Period depends on mass and stiffness, not amplitude. \( T = 2\pi\sqrt{m/k} \). Pull farther or harder — same period, just more energy and higher max speed.
Maximum speed occurs at equilibrium. \( v_{\text{max}} = A\sqrt{k/m} \) — it scales directly with amplitude.

Key
\( k \) = spring constant (N/m)
\( x \) = displacement from equilibrium
\( A \) = amplitude
\( T \) = period (s) · \( f \) = frequency (Hz) All Your Equations:

\( F_s = -kx \)
\( U_s = \frac{1}{2}kx^2 \)
\( E_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \)
\( v_{\text{max}} = A\sqrt{k/m} \)
\( T = 2\pi\sqrt{m/k} \)
\( f = 1/T \)