An Oscillating Spring
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 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 analogous 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.
Analogous to a pendulum, one can apply the conservation of mechanical energy to the mass-spring 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
½ mv 2 = ½ kA 2 v = (kA 2 /m) 1/2 = (k/m) 1/2 A

The following equations will be accepted without proof. x and y components of displacement expressed as a function of time:
 x = Asin(ωt)
 y = Acos(ωt) x and y components of velocity expressed as a function of time:

v x
= ωAcos(ωt)

v y
= -
ωAsin(ωt) x and y components of acceleration expressed as a function of time:

a x
= -
ω2Asin(ωt)

a y
= -
ω 2 Acos(ωt)
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:

When the mass is large, the motion will be slow corresponding to a large period and a small frequency.

When the spring constant, k, is large this results in a stiff spring and the periodic motion will have small period and a large frequency.

The period and frequency are independent of amplitude and the total mechanical energy.

The period is dependent upon the mass and the spring constant.

Oscillating Spring Problems
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
The amplitude, A, is 0.050 m because the mass-spring system was given a
maximum displacement of 0.050 m.
(b) period
m = 5.0 kg A = 0.050 m Δx = 0.10 m
F r
= kx
k = F r
/Δx = F w
/Δx = mg/Δx = 5.0 kg*9.80 m/s 2 /0.10 m = 490 N/m
T = 2π(m/k) 1/2 = 2π(5.0 kg/490 N/m) 1/2 = 0.63 s
(c) frequency
f = 1/T = T -1 = (0.63 s) -1 = 1.6 /s = 1.6 Hz

2) A 100. g mass hangs from a vertical spring. When it is pulled 10.0 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?
m = 100. g A = 10.
cm * 1 m/10 2 cm = 0.10 m
T = 2π(m/k) 1/2 k = 4π 2 * 100.
G * 1 kg/10 3 g/(2.0 s) 2 = 0.99 N/m
(KE + PE e
)
A
= (KE + PE e
) eq pos
0 + ½ kx 2 = ½mv 2 + 0
½*0.99 N/m*(0.10
m) 2 = ½*100.
g*1 kg/10 3 g*v 2
v = 0.31 m/s
(b) What is the acceleration when it is 5.0 cm below the equilibrium position?
F = ma = -kx
a = -0.99 N/m * 5.0
cm * 1 m/10 2 cm/(100.
g * 1 kg/10 3 g) = -.50 m/s 2
(c) How much will the spring shorten if the mass is removed?
F = kx
x = F/k = F w
/k = mg/k = 100.
g * 1kg/10 3 g * 9.80 m/s 2 /0.99 N/m = 0.99 m

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 the:
(a) force constant of the spring
x = 0.0400 m A = 0.300 m m = 0.150 kg
F = kx k = F/x = F w
/x = mg/x = 0.150 kg*9.80 m/s 2 /0.0400 m = 36.8 N/m
(b) period of the motion
T = 2π(m/k) 1/2 = T = 2π(0.150 kg/36.8 N/m) 1/2 = 0.401 s
(c) displacement of the mass as a function of time y = Acos(ωt)
ω = 2πf = 2π/T = 2*3.14/0.401 s = 15.7 s -1
y = 0.300cos(15.7t)*m
(d) velocity of the mass as a function of time
v y
= ωAsin(ωt)
v y
= -15.7*0.300sin(15.7t) = -4.71sin(15.7t)
(e) acceleration of the mass as a function of time
a y
= ω 2 Acos(ωt)
a y
= -(15.7) 2 * 0.300cos(15.7t) = -73.9cos(15.7t)

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
f = 3.0 s -1 A = 16 cm v = Aωcos(ωt + Φ)
v = 16 cm * 1 m/10 2 cm *
2π
* 3.0 s -1 = 3.0 m/s a = Aω 2 sin(ωt + Φ) a = -16 cm * 1 m/10 2 cm *
(2π
* 3.0 s -1 ) 2 = -57 m/s 2
Keep in mind that both the cos(ωt + Φ) and the sin(ωt + Φ) vary from a minimum
of 0 to a maximum of 1.
(b) velocity and acceleration when the mass has a displacement of 7.0 cm x = Asin(ωt + Φ)
7.0 cm = 16 cm* sinθ
sin -1 θ = 7.0
cm/16 cm
 = 26° v = Aωcosθ =
16 cm * 1 m/10 2 cm *
2π
* 3.0 s -1 * cos
26°
= 2.7 m/s a = Aω 2 sin(ωt + Φ)
a = -16 cm * 1 m/10 2 cm *
(2π
* 3.0 s -1 ) 2 * sin
26°
= -25 m/s 2
(c) time needed to move from the equilibrium position to a displacement of 13 cm x = Asin(ωt + Φ)
sin -1 θ = x/A = 13cm/16 cm
θ = 0.948
θ = ωt t = 0.948/2/π/3.0 s -1 = 0.050 s