Lesson 8, Topic 2
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# Vertical Oscillations of a Spring Mass System

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Let’s say we have rigid support and a string hanging from the support. Let’s say we then attach a mass, there will be a small elongation, e, in the spring.

At this point the spring-mass system is in equilibrium, therefore, all the forces acting will cancel out each other, i.e the net force = 0.

Fnet = 0

The two forces are the restoring force acting upwards and mg acting downwards and their net force is zero.

âˆ´ F + mg = 0

F = -mg

We know F = –ke

ke + mg = 0

âˆ´ –ke = -mg

âˆ´ ke = mg ………. *

From the equilibrium position, if a downward force is applied, the attached mass will be displaced vertically through a distance of x and the system will oscillate with an acceleration, a.

Restoring force, F’ = -k(e + x) ……..(1)

The net force towards centre = Fnet

Fnet = F’ + mg

Fnet = -k(e + x) + mg …..(2)

Note: It is not equal to 0 because it is not in equilibrium because the spring is oscillating with an acceleration, a.

Open the brackets in equation 2

Fnet = -ke – kx + mg

Since ke = mg, then

Fnet = -mg – kx + mg

But Fnet = ma

âˆ´ ma = -kx

a = $$\normalsize \frac{-k}{m} \scriptsize x$$ ……….. (3)

Therefore we can say;

$$\scriptsize a \propto -x$$

When acceleration is proportional to the displacement and in the opposite direction of displacement the oscillations are simple harmonic.

a = -Ï‰2x

a = $$\normalsize \frac{-k}{m} \scriptsize x$$ ……….. (3)

we can see that

$$\scriptsize \omega^2 = \normalsize \frac{k}{m} \\ \scriptsize \: or \: \omega = \normalsize \sqrt{\normalsize \frac{k}{m}}$$

The period of motion, T = $$\frac{2 \pi}{\omega}$$

Let’s substitute Ï‰ into the equation T = $$\frac{2 \pi}{\omega}$$

Ï‰ = $$\sqrt {\frac{k}{m} }$$

âˆ´ T = $$\frac{2 \pi}{\sqrt {\large \frac{k}{m}}}$$

T = $$\scriptsize 2 \pi\: \times \: \sqrt {\normalsize \frac{m}{k}}$$

âˆ´ T = $$\scriptsize 2 \pi \sqrt {\normalsize \frac{m}{k}}$$

but T = $$\frac{1}{f}$$

âˆ´ f = $$\frac{1}{2 \pi} \sqrt {\normalsize \frac{k}{m}}$$

### Example 1:

A horizontal spring (k = 300 N/m) has a mass of 0.85 kg attached to it and is undergoing simple harmonic motion. Calculate the
(a) period
(b) frequency
(c) angular frequency

Solution:

(a) T = $$\scriptsize 2 \pi \sqrt {\normalsize \frac{m}{k}}$$

T = $$\scriptsize 2 \pi \sqrt {\normalsize \frac{0.85}{300}}$$

T = $$\scriptsize 2 \pi \sqrt {0.00283}$$

T = $$\scriptsize 2 \pi \: \times \: 0.053$$

T = 0.333s

(b) f = $$\frac{1}{T}$$

f = $$\frac{1}{0.333}$$

f = 3Hz

(c) Ï‰ = 2Ï€f

Ï‰ = 2 x Ï€ x 3

### Example 2:

A 0.95kg mass vibrates according to the equation x = 0.76cos(7.84t). Calculate the
(a) amplitude
(b) frequency
(c) period
(d) spring constant

Solution:

(a) compare equation to x = A cos (Ï‰t)

amplitude = 0.76

âˆ´ $$\scriptsize -0.76 \leq x \leq +0.76$$

(b) angular speed Ï‰ = 7.84 rad/s

Ï‰ = 2Ï€f

f = $$\frac{\omega}{2 \pi}$$

f = $$\frac{7.84}{2 \pi}$$

f = 1.25 Hz

(c) T = $$\frac{1}{f}$$

T = $$\frac{1}{1.25}$$

T = 0.8s

(d) To find spring constant k use the formula;

T = $$\scriptsize 2 \pi \sqrt {\normalsize \frac{m}{k}}$$

make k the subject,

square both sides

T2 = $$\scriptsize (2 \pi)^2 \normalsize \frac{m}{k}$$

cross multiply

kT2 = $$\scriptsize (2 \pi)^2 \: \times \: m$$

k = $$\frac {(2 \pi)^2 \: \times \: m}{T^2}$$

k = $$\frac{(6.283)^2 \: \times \: 0.95}{0.8^2}$$

k = $$\frac{39.5 \: \times \: 0.95}{0.64}$$

k = $$\frac{37.525}{0.64}$$

k = 58.63 N/m

error: