MechanicalVibrationseasyFE ExamCoursework

Natural Frequency of a Spring-Mass System

A mass of $m = 5.0\ \text{kg}$ is suspended from a spring of stiffness $k = 2000\ \text{N/m}$ .

Find the undamped natural frequency in hertz.

Given

m	5 kg	Mass
k	2000 N/m	Spring stiffness

Hint 1

ω_n = √(k/m) gives the natural frequency — but in what units?

Hint 2

That formula returns radians per second. The question asks for hertz.

Hint 3

f = ω/(2π).

Worked solution — try the problem first

Natural angular frequency:

$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{2000}{5.0}} = \sqrt{400} = 20.0\ \text{rad/s}$

Convert to hertz. The question asks for cycles per second, not radians per second:

$f_n = \frac{\omega_n}{2\pi} = \frac{20.0}{2\pi} = 3.183\ \text{Hz}$

$\boxed{f_n = 3.18\ \text{Hz}}$

Note. Gravity does not appear. Hanging the mass shifts the equilibrium position by $mg/k$ , but the oscillation frequency about that new equilibrium is unchanged — a vertical spring-mass system and a horizontal one have the same natural frequency.

Concepts:Natural frequencySimple harmonic motionFree vibration

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