ใฐ๏ธSprings & Hooke's Law
Hooke's Law
Hooke's Law describes the force exerted by an ideal spring:
F_spring = โkx
Where: - k = spring constant (N/m) โ stiffness of the spring - x = displacement from natural length - Negative sign: force opposes displacement (restoring force)
A stiffer spring has a larger k. Springs in parallel have k_eff = kโ + kโ. Springs in series: 1/k_eff = 1/kโ + 1/kโ.
Period of a Spring-Mass System
For a mass on a spring, the period is:
T = 2ฯโ(m/k)
Key observations: - More mass โ longer period (heavier objects oscillate more slowly) - Stiffer spring (larger k) โ shorter period (stiffer spring oscillates faster) - Amplitude doesn't affect period! (A feature of SHM)
The frequency: f = (1/2ฯ)โ(k/m)
Spring-Mass Oscillator
Adjust the mass and spring constant. Notice how they affect the period but amplitude doesn't.
โ๏ธ Worked Example
Problem: A 0.5 kg mass hangs from a spring with k = 200 N/m. Find the period and frequency of oscillation.
๐ Key Equations
Hooke's Law and spring-mass system
F_s = -kxT_spring = 2ฯโ((m)/(k))f = (1)/(2ฯ)โ((k)/(m))โ ๏ธ Common Mistakes
Misconception: Forgetting the negative sign in Hooke's Law, writing F = kx instead of F = โkx.
โ Correct thinking: The negative sign is essential โ it means the spring force is always directed opposite to the displacement (a restoring force). Without it, the force would push the object further away from equilibrium.
Why: This restoring force is exactly what produces oscillation. A positive force for positive displacement would cause exponential runaway, not SHM.
Misconception: Thinking a stiffer spring (higher k) means a longer period because it's "harder to compress."
โ Correct thinking: A stiffer spring (larger k) gives a SHORTER period: T = 2ฯโ(m/k). Higher k pulls the mass back more forcefully, so it oscillates faster.
Why: Think of it this way: k is in the denominator of m/k, so increasing k decreases T.
Misconception: Applying T = 2ฯโ(m/k) to a spring that hasn't been set into oscillation (just statically stretched).
โ Correct thinking: T = 2ฯโ(m/k) describes the oscillation period when the spring is disturbed from equilibrium. Static equilibrium (object hanging at rest) uses F = kx = mg to find the stretch, not T.
Why: These are two different problems: static equilibrium finds the rest position, while the period formula describes the dynamics of oscillation around that position.
๐ Practice Problems
Try these problems. Check your answer when ready.
A spring with k = 50 N/m is stretched 0.1 m from equilibrium. What is the restoring force?
A 1 kg mass is attached to a spring with k = 400 N/m. Find the period and frequency of oscillation.
A spring-mass system has a period of 1.2 s with a 0.3 kg mass. What is the spring constant k?
T = 2ฯโ((m)/(k))Two springs in parallel (kโ = 40 N/m, kโ = 60 N/m) support a 2 kg mass. What is the period of oscillation?
A spring-mass system on a frictionless surface has k = 200 N/m and m = 0.8 kg. It is released from rest at x = 0.15 m. Find: (a) period, (b) maximum speed, (c) speed at x = 0.1 m.
v = ฯโ(Aยฒ - xยฒ)Finished reading through this lesson?