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Mass between two springs, period

Short answer

For a simple mass-spring, m d²x/dt² = –kx so T = 2π√(m/k). With mass M between two springs K1, K2 on a frictionless table, displacing x stretches one and compresses the other, net restoring force = –(K1+K2)x. Thus T = 2π√(M/(K1+K2)).

Explain the theory of a simple mass spring system. Develop the relation for time period and frequency of two springs having spring constants K1 and K2 supporting a mass 'M' between them on a frictionless horizontal table.

Simple mass-spring system

A mass m attached to a spring of constant k on a frictionless surface. When displaced by x, the spring exerts the restoring force F = −kx (Hooke's law).

m d²x/dt² = −kx   →   d²x/dt² + (k/m) x = 0
x = a sin(ωt + φ) ,  ω = √(k/m) ,  T = 2π√(m/k) ,  f = (1/2π)√(k/m)

The motion is SHM; the period depends on m and k only.

Mass between two springs

A mass M rests on a smooth horizontal table between two fixed walls, joined to them by springs of constants K1 and K2.

 |W|~~~K1~~~[ M ]~~~K2~~~|W|
              <-x->
  1. At equilibrium both springs are unstretched (or equally pre-stretched). Displace M by x to the right.
  2. Spring 1 is stretched by x and pulls the mass back with force K1 x. Spring 2 is compressed by x and pushes the mass back with force K2 x. Both act in the same direction (towards equilibrium).
  3. Net restoring force:
F = −(K1 + K2) x
  1. Equation of motion: M d²x/dt² = −(K1 + K2) x, i.e. d²x/dt² + [(K1 + K2)/M] x = 0.
  2. This is SHM with equivalent spring constant K = K1 + K2:
ω = √((K1 + K2)/M)
T = 2π √(M/(K1 + K2))
f = (1/2π) √((K1 + K2)/M)

The springs in this arrangement behave as springs in parallel, so the effective stiffness is higher and the period is shorter than for either spring alone.

✓ Answer checked by Claude Opus 5.5 (high effort)

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