Table of Contents
An Oscillating Spring
Directions 

An oscillating spring is an example of SHM (Simple Harmonic Motion) which results when the force is directly 
proportional to the displacement and is always directed toward the equilibrium position. As a result of the restoring force 
changing, the acceleration is also changing, and the kinematics equations for constant acceleration do not apply. 
Using the example of a mass attached to a vertical spring as shown in the animation gives the result: 
F = ma = kx

a = k/m • x

where k is the spring constant with units of N/m (newtons/meter), m is the mass of the object with units of kg, and x is 
the displacement from the equilibrium with units of m (meters). The negative sign results from the restoring force always 
opposing the motion and being directed toward the equilibrium position. The second equation demonstrates that the 
acceleration at each instant is directly proportional to the negative displacement at that instant. 

This situation is analagous to that of a simple pendulum. When x is a maximum at the amplitude, the restoring force, Fr, 
and the resulting acceleration is also a maximum. At this point, the velocity is zero. When x is zero at the equilibrium 
position, both Fr and a are zero with the velocity a maximum. 
Analagous to a pendulum, one can apply the conservation of mechanical energy to the massspring system. 
ΔE = 0

(KE + PE_{e})_{eq} = (KE + PE_{e})_{A}

where eq signifies the equilibrium position and A is the amplitude. 
KE + 0 = 0 + PE_{e}

1/2mv^{2} = 1/2kA^{2}

v = (kA^{2}/m)^{1/2} = (k/m)^{1/2}A

The following equations will be accepted without proof. 









Modifying the x component of displacement yields: 
x = Asin((k/m)^{1/2} • t)

The quantity in parenthesis is the angle measured in radians. Because x is a function of time, it is said to be a sinusoidal 
function. The sin function repeats itself when the quantity (k/m)^{1/2} • t increases by 2π. The period of the oscillating 
spring is given by: 
T = 2π(m/k)^{1/2}

Examining this equation^{} leads to four important considerations: 

large frequency. 


1) A 5.0 kg mass is attached to a vertical spring and stretches the spring 0.10 m. The spring is then pulled down 
0.050 m and released. Determine the: 
(a) amplitude 
(b) period 
(c) frequency 
2) A 100. g mass hangs from a vertical spring. When it is pulled 10. cm below its equilibrium position and released, it 
oscillates with a period of 2.0 s. 
(a) What is the maximum velocity of the mass? 
(b) What is the acceleration when it is 5.0 cm below the equilibrium position? 
(c) How much will the spring shorten if the mass is removed? 
3) When a 0.150 kg mass is attached to a vertical spring, it stretches by 0.0400 m. The spring is then pulled down 
0.300 m below the equilibrium position and released. Determine: 
(a) force constant of the spring 
(b) period of the motion 
(c) mass’s displacement as a function of time 
(d) mass’s velocity as a function of time 
(e) mass’s acceleration as a function of time 
4) A mass attached to a vertical spring is vibrating with a frequency of 3.0 s^{1} and has an amplitude of 16 cm. Determine 
the: 
(a) maximum velocity and the maximum acceleration 
(b) velocity and acceleration when the mass has a displacement of 7.0 cm 
(c) time needed to move from the equilibrium position to a displacement of 13 cm 



