In previous lessons we spoke about simple harmonic motion, its definition, systems that engage in simple harmonic motion and formulas involved. Now we’ll be looking at calculating Simple harmonic motion questions.

Example # 1

S.H.M

In the diagram above a particle is moving with S.H.M with a period of 24 seconds between two points A and B. Find the time taken to travel from:
a) A to B
b) O to B
c) O to C
d) D to E

Answers:
Note that the period is the time taken to complete one revolution in this case the period is the time taken for the particle to travel from A to B and then to A again (24 seconds as stated in the question above).

a) A to B
A to B is half a period which would be = 12 second

If you want to work this question in a more detailed way, you can do this:

Distance from A to B to A again = 16m
Distance from A to B = 8m
Period (A to B to A again) = 24 seconds
Time (t) from A to B =???

b) O to B
O to B is quarter of the period so the answer would be = 6 sec

Or you can follow the above pattern:

Distance from A to B to A again = 16m
Distance from O to B = 4m
Period (A to B to A again) = 24 seconds
Time (t) from O to B =???

c) O to C
This is one eight of the period which would be = 3 seconds

Or by detailed calculation:

Distance from A to B to A again = 16m
Distance from O to C = 2m
Period (A to B to A again) = 24 seconds
Time (t) from O to C =???

d) D to E
Note with this question, if you observe the diagram carefully you will see that you weren’t given the distance between D to E directly. However this can easily be found by subtracting the distance O to D from the distance O to E and this would give you the distance from D to E.

Distance from A to B to A again = 16m
Distance from D to E = 3.5 – 3 = 0.5m
Period (A to B to A again) = 24 seconds
Time (t) from D to E =???

What is Moments?

By definition Moments, also known as torque, is the turning effect of a force. Moments can either be in a clockwise or anti-clockwise direction. The unit of moments is the Newton Meter (Nm).

Formula for Moments of a force:

      Moments (torque) = Force × Perpendicular distance from the pivot

How to calculate Moments?

Example 1
Find the total moments of the system below:

Diagram Showing Moments

Step 1

Identify which force in the system is moving in the clockwise direction and which is moving in an anticlockwise direction. The 50N force is the one moving in the clockwise direction (the same direction a clock’s pointer would move) while the 80N is moving in the opposite direction (anti-clockwise direction).

Showing Clockwise and Anti-clockwise Moments

Step 2

Use the formula given above to calculate the clockwise and anti-clockwise moments separately.

          Moments = Force × Perpendicular distance

          Clockwise Moments = Force × Perpendicular Distance
                                          = 50N × 6m
                                          = 300Nm

          Anti-Clockwise Moments = Force × Perpendicular Distance
                                                  = 80N × 4m
                                                  = 320Nm

Step 3

Now that you have calculated both clockwise and anti-clockwise moments you can now find the total moments of the system. This is found by subtracting the smaller moments from the larger, in this case the smaller of the two is the clockwise while the larger is the anti-clockwise.

         Total Moments = 320Nm – 300Nm
                                 = 20Nm in the anti-clockwise direction

 Note: Anti-clockwise direction is written at the end because it is larger.

Projectile Motion

What is the projectile motion of a body?
- A body in projectile motion will move in two dimensions simultaneously. Bodies that are in projectile motion are resolved(divided) into two components. A vertical component(Vy) and an horizontal component(Vx). These components however are dealt with simultaneously.


Please refer to the diagram below showing projectile motion.

Projectile motion

Formulas associated with projectile motion include:

• Vy = VsinA
• Vx = VcosA
• V = Sqrt [Vy^2 + Vx^2]
• TanA = Vy / Vx => A = Tan^-1 A (Vy / Vx)

Where A is used as angle instead of theta in diagram

Example calculation involving projectile motion:

1. A body is projected with a velocity of 200m/s at and angle of 30 °C to the horizontal.

• Calculate the time taken to reach its maximum height.
• Calculate its velocity after 16 seconds.

Answers:

• Finding the time for maximum height.

angle A = 30 °C
          V = 200m/s
Acceleration(a) = -10 m/s²
Vy = 0 m/s

Therefore Uy = VsinA                                                [ Note: Uy represents initial velocity]

                  Uy = 200 x Sin 30

                  Uy = 100 m/s

                  

                  Vy = Uy = at

                  Vy - Uy = at

                        a

                   0 - 100 = t

                      - 10

                      10 sec = t

Therefore Maximum Time (t) = 10 seconds

• Finding the velocity after 16 seconds.

                  Vx = V cos A 
                  Vx = 200 x Cos 30
                  Vx = 173.2 m/s

                  Vy = Uy + at
                  Vy = 100 - 10 (16)
                  Vy = -60 m/s

Therefore V =  √[Vy^2 + Vx^2]
                      =  √[(-60)^2 + (173.2)^2]
                      =  √33598
                      =  183.30 m/s