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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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Below we show that the motion of the bob of a simple pendulum is simple harmonic.

If a bob of mass, m, of a simple pendulum is slightly displaced through a small angle and allowed to oscillate about O.

pendulum 2 e1657322949635
Fig. 1 – Theory of Simple Pendulum.

Suppose it is at A, at any instance in time during oscillations and θ is the angle POA, subtended by the string with the vertical.

mg is the force acting on the bob at point A in a vertically downward direction.

At A, the bob is at its extreme position. its velocity is zero, and its component mg cos θ is balanced by the tension in the string which is clear from fig. 2 below.

Screenshot 2022 07 08 at 22.46.47
Fig. 2

Hence the equation becomes: T − mgcosθ = 0

The force that tends to move it towards the centre O, in the direction opposite to increasing θ = -mg sinθ

i.e, F = -mgsinθ

Screenshot 2022 07 08 at 22.46.18
Fig. 3

and F = ma

∴ -mg sinθ = ma ………**

Period (T) is the time it takes the pendulum to complete one cycle. i.e to swing from A to O to B, and from B to O back to A.

For the period, T, of oscillation;

The angle of displacement, θ is small, such that the arc OA forms an approximately straight line if OA = x.

When θ is small, sin θ = θ in radians.

Also from fig. 1

θ = \( \frac{x}{l}\)

∴ from equation **

-mg sinθ = ma, becomes

\(\scriptsize -mg \normalsize \frac{x}{l}\) = ma

let’s make a the subject of the formula

ma = \( \scriptsize -mg \normalsize \frac{x}{l}\)

a = \( \scriptsize -\normalsize \frac{\not{m}}{\not{m}} \scriptsize g \normalsize \frac{x}{l}\)

a = \( \frac{-g}{l}\scriptsize x\)

a = -ω2x

where ω2 = \( \frac{g}{l} \) or ω = \(\sqrt { \frac{g}{l}}\)

angular velocity (ω) of oscillating pendulum

Since the acceleration a of the oscillating mass is proportional to the distance, x from O, we can say that the motion is simple harmonic.

Since T = \(\frac{2 \pi}{ω} \)

T \( \scriptsize = 2 \pi \sqrt { \normalsize \frac{l}{g}}\)

The formula only works for the oscillations through small angles, as it was something we assumed in the process of proof.

Example 1:

What is the period and frequency of a simple pendulum that has a length of 1.10m?
g = 9.8 m/s2

Solution:

Use the formula;

l = 1.10m

g = 9.8 m/s2

T \( \scriptsize = 2 \pi \sqrt { \normalsize \frac{l}{g}}\)

T \( \scriptsize = 2 \pi \sqrt { \normalsize \frac{1.10m}{9.81}}\)

T \( \scriptsize = 2 \pi \sqrt { 0.112s}\)

T = 2.10s

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

f = \( \frac{1}{2.10s} \)

f = 0.48 Hz

Example 2:

A pendulum with a mass of 0.25 kg has a period of 1.15 seconds. If the initial displacement of the pendulum is 20º, what is the length of the pendulum?

Solution:

To find the length of the pendulum use the formula;

T \( \scriptsize = 2 \pi \sqrt { \normalsize \frac{l}{g}}\)

male l the subject of the formula

square both sides

T2 \( \scriptsize = \left(2 \pi \sqrt { \normalsize \frac{l}{g}} \right)^2\)

open the brackets

T2 \( \scriptsize = 4 \pi^2 \normalsize \frac{l}{g}\)

\( \scriptsize T^2g = 4 \pi^2 l \)

\( \scriptsize l = \normalsize \frac{T^2g}{4 \pi^2} \)

\( \scriptsize l = \normalsize \frac{1.15^2 \: \times \: 9.81 }{4 \pi^2} \)

\( \scriptsize l = \scriptsize 0.33 \: m \)

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