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SS2: PHYSICS - 1ST TERM

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  1. Scalars & Vectors | Week 1
    5 Topics
    |
    1 Quiz
  2. Equations of Motion | Week 2
    3 Topics
    |
    1 Quiz
  3. Projectile | Week 3
    5 Topics
  4. Equilibrium of Forces I | Week 4
    4 Topics
  5. Equilibrium of Forces II | Week 5
    4 Topics
  6. Stability of a Body | Week 6
    4 Topics
    |
    1 Quiz
  7. Simple Harmonic Motion (SHM) | Week 7
    4 Topics
  8. Speed, Velocity & Acceleration & Energy of Simple Harmonic Motion | Week 8
    5 Topics
    |
    1 Quiz
  9. Linear Momentum | Week 9
    6 Topics
    |
    1 Quiz
  10. Mechanical Energy & Machines | Week 10
    2 Topics
    |
    1 Quiz



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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.

vert oscillations

∴ F + mg = 0

F = -mg

We know F = –ke

ke + mg = 0

∴ –ke = -mg

∴ 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.

vert oscillations 3

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

∴ 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.

Comparing equation (3) with SHM equation,

a = -ω2x

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 ω into the equation T = \( \frac{2 \pi}{\omega} \)

ω = \( \sqrt {\frac{k}{m} }\)

∴ T = \( \frac{2 \pi}{\sqrt {\large \frac{k}{m}}} \)

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

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

but T = \( \frac{1}{f} \)

∴ 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) ω = 2πf

ω = 2 x π x 3

= 18.85 rad/s

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 (ωt)

amplitude = 0.76

∴ \( \scriptsize -0.76 \leq x \leq +0.76 \)

shm question

(b) angular speed ω = 7.84 rad/s

ω = 2πf

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

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