Example 1
k = 200 N/m, m = 0.5 kg.
Spring Oscillation Frequency Calculator
Enter the spring constant and the mass. You get f and the period T = 2π√(m/k). k = 200 N/m and m = 0.5 kg is f ≈ 3.18 Hz and T ≈ 0.314 s. Amplitude is not in the formula.
Force at a given x: F = k·x. Energy: Eₛ = ½kx². T from this f: T = 1/f. ω: angular velocity.
Result
Spring constant k (N/m) and Mass m (kg). The result shows up here.
How it works
In an ideal mass-spring system the frequency is f = (1/(2π))√(k/m) and the period is T = 2π√(m/k). k = 200 N/m and m = 0.5 kg give √(k/m) = 20, so f = 20/(2π) ≈ 3.18 Hz and T ≈ 0.314 s. Doubling the mass to 1 kg drops f to about 2.25 Hz, not to half.
The k field is always newtons per meter. The m field is kilograms. Both must be positive. There is no amplitude field: 5 cm and 1 cm give the same f while the spring stays in Hooke’s linear range.
A stiffer spring (larger k) raises f. A larger mass lowers f like 1/√m. g = 9.81 does not enter the formula. A vertical spring with sag mg/k only shifts the equilibrium, it does not change f. A pendulum computes T from L and g, a different card.
k = 50 N/m and the same m = 0.5 kg give √(k/m) = 10, so f ≈ 1.59 Hz. k = 800 and m = 1 kg is f ≈ 4.50 Hz. k = 1000 and m = 0.1 kg is √(k/m) = 100 and f ≈ 15.9 Hz. A comma and a period both parse.
Type k and m, for example 200 and 0.5, then Calculate. The result is f and T. Check: open oscillation period, type 3.18, and you come back near 0.314 s. Force at a given x lives on Hooke’s law. Energy ½kx² is next door.
ω = √(k/m) = 2πf. At k = 200 and m = 0.5 that is 20 rad/s. You can drop that ω into angular velocity, or into rotational Eₖ if you are building an analogy, but the spring energy here is ½kx², not ½Iω².
How to use
- Type the spring constant k in N/m. The calculator example is 200. You get k from sag under weight (kx = mg) or from a datasheet.
- Type mass m in kilograms, for example 0.5. That is the oscillating mass, not the coil mass in the school model.
- Click Calculate. 200 N/m and 0.5 kg give f ≈ 3.18 Hz and T ≈ 0.314 s. Mass 1 kg at that k is about 2.25 Hz.
- You do not type amplitude. 5 cm and 1 cm give the same f while F = kx stays linear.
- You can check T = 1/f on oscillation period. Force and energy live on Hooke’s law and elastic energy.
Formula
f = (1/(2π)) √(k/m)
T = 2π √(m/k) = 1/f
Letters in f = (1/2π)√(k/m)
A mass on a spring: f = (1/2π)√(k/m). k = 200 N/m and m = 0.5 kg give √(k/m) = 20, so f ≈ 3.18 Hz and T ≈ 0.314 s.
- f
- Frequency in hertz. 20/(2π) ≈ 3.18 Hz at 200 N/m and 0.5 kg; doubling the mass to 1 kg drops f to ≈ 2.25 Hz.
- k
- Spring constant in N/m. 200 at 0.5 kg sets that 3.18 Hz; amplitude has no field and is not in the formula.
- m
- Mass in kilograms. 0.5 kg at k = 200 is T ≈ 0.314 s; larger m lowers f like a square root, not linearly.
- T
- Period 2π√(m/k) = 1/f. In this example ≈ 0.314 s, a companion result to f, not a separate field.
Real-life examples
Example 2
k = 200 N/m, m = 0.2 kg.
Example 3
k = 50 N/m, m = 0.5 kg.
Example 4
k = 800 N/m, m = 1 kg.
Example 5
k = 20000 N/m, m = 300 kg.
Example 6
k = 500 N/m, m = 0.05 kg.
Example 7
k = 100 N/m, m = 0.25 kg.
Example 8
k = 200 N/m, m = 2 kg.
Example 9
k = 150 N/m, m = 0.4 kg.
Example 10
k = 1000 N/m, m = 0.1 kg.
Ways to use this calculator
- Lab: from k and m you read the spring f and T, then compare with a stopwatch.
- A bridge to F = kx, to ½kx², and to the T ↔ f pair on the neighboring pages.
Frequently asked questions
What T and f at k = 200 N/m and m = 0.5 kg?
f ≈ 3.18 Hz, T ≈ 0.314 s. Mass 1 kg at that k drops f to ≈ 2.25 Hz, not to 1.59 Hz.
Does a 5 cm amplitude change f?
No. The formula has no x. 5 cm and 1 cm both give 3.18 Hz while the spring stays linear.
Where does k come from?
From sag under weight: kx = mg, so k = mg/x. Or from a datasheet. Force at x is on Hooke’s law.
Does g = 9.81 enter f?
No. A pendulum uses √(L/g). Here it is √(k/m). Vertical sag mg/k is a new zero, not a new f.
How does this link to energy?
Eₛ = ½kx² on elastic energy. In oscillation that energy turns into Eₖ and back.
What about ω?
ω = √(k/m) = 2πf. At 200 and 0.5 that is 20 rad/s. See angular velocity too.
Can I type 0.5 kg with a comma?
Yes. 0,5 and 0.5 mean the same mass [kg]. Then f = (1/2π)√(k/m).
What f at k = 50 N/m and m = 0.5 kg?
√(k/m) = 10, so f ≈ 1.59 Hz and T ≈ 0.628 s. That is not the F = kx card.
How does f fall when I double m?
Like 1/√m. Doubling mass multiplies f by 1/√2 ≈ 0.707, not by 1/2.
Can I type k in N/cm?
The field wants N/m. 2 N/cm is 200 N/m, the same 3.18 Hz example at 0.5 kg.
Knowledge sources
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.