Simple Harmonic Motion (SHM) involves the back-and-forth movement of an object around a central point. The energy of SHM is divided into two main components:

1. Kinetic Energy (KE): This is the energy associated with the motion of the object. In SHM, when the object is at its maximum displacement, it has maximum kinetic energy. At the equilibrium position, kinetic energy is minimum.

2. Potential Energy (PE): The potential energy in SHM is related to the displacement of the object from its equilibrium position. It is maximal when the object is farthest from the equilibrium and minimal (or zero) when the object is at the equilibrium position.

In mathematical terms, the total mechanical energy (E) of an object in SHM is the sum of its kinetic and potential energy:

\[E = KE + PE\]

From the law of conservation of energy; $$ K.E = P.E $$ $$ \frac{1}{2}mv²{_m} = mgh $$ Where, V = max. velocity

The energy stored in S.H.M can be derived using the motion of a mass suspended from the end of a spiral spring. As the mass is displaced, it experiences an elastic restoring force which restores the spring back to its mean position. From Hooke's law; $$ F = -Kx $$ Where; y = distance of the spring from its mean position
K = force constant
The force is negative to show it acts in a direction opposite the motion.

The workdone by the spring is given by: $$ W = average force × distance $$ $$ W = \frac{1}{2}kx × x $$ $$ W = \frac{1}{2}kx² $$ The maximum total energy stored in the spring is given by: $$ W = \frac{1}{2}kA² $$ Where A = Amplitude (maximum displacement)

At any stage of oscillation, the total energy of the spring is given by: $$ W = K.E + P.E $$ $$ W = \frac{1}{2}kA² = \frac{1}{2}mv² + \frac{1}{2}kx² $$ $$ \frac{1}{2}kA² - \frac{1}{2}kx² = \frac{1}{2}mv² $$ $$ k(A²-x²) = mv² $$ $$ v = {\sqrt{\frac{k}{m}(A²-x²)}} $$ But, $$ v = \omega {\sqrt{A²-x²}} $$ Comparing both equations; $$ \omega = \sqrt{\frac{k}{m}} $$ $$ T = \frac{2 \pi}{ \omega} $$ $$ T = 2\pi \sqrt{\frac{m}{k}} $$ Assuming the mass of the spring is negligible.
Note: Hence \( T \alpha \sqrt{m} \)

Example 1: A 5kg mass is hung from the end of a coil spring. If the system is made to vibrate with simple harmonic motion of period 2 secs, find the energy of the motion if the amplitude of vibration is 3cm.

To use the formula, first find the value for force constant, k $$ \omega = {\sqrt{\frac{k}{m}}} $$ But, $$ \omega = \frac{2 \pi}{T} $$ $$ \omega = \frac{2 × 3.142}{2} $$ $$ \ omega = \pi rad/s $$ $$ \omega² = \frac{k}{m} $$ $$ k = \omega² × m $$ $$ k = π² × 5 $$ $$ k = 5π² = 49.35N/m $$ $$ W = \frac{1}{2}kA² $$ $$ W = \frac{1}{2}×49.35 × 0.03² $$ $$ W = 0.022J $$

Example 2: A body of mass 200g suspended from the end of a spiral spring stretches the spring by 4cm. When the body is set in motion, it attains a maximum displacement of 5cm . Calculate:

1. The force constant of the spring
2. The period of oscillation
3. The frequency of oscillation
4. The angular speed of the body
5. Total energy stored in the spring
6. Maximum velocity and acceleration attained by the spring.

1. To solve for force constant;
   From Hooke's law: $$ F = ke $$ $$ mg = ke $$ $$ k = \frac{0.2 × 10}{0.04} $$ $$ k = 200N/m $$
2. The period is calculated as thus: $$ T = 2 \pi \sqrt{\frac{m}{k}} $$ $$ T = 2 \pi \sqrt{\frac{0.2}{200}} $$ $$ T = 0.199s $$
3. The frequency is calculated as : $$ frequency = \frac{1}{T} $$ $$ f = \frac{1}{0.199} $$ $$ f = 5.03Hz $$
4. The angular velocity can be calculated as: $$ \omega = \sqrt{\frac{k}{m}} $$ $$ \omega = \sqrt{\frac{200}{0.2}} $$ $$ \omega = 31.62rad/s $$
5. The total energy stored in the spring is given as: $$ W = \frac{1}{2}KA² $$ $$ W = \frac{1}{2}× 200 × 0.05² $$ $$ W = 0.25J $$
6. The maximum velocity, V_max attained at mid point is given by: $$v_{max} = \omega A $$ $$ v_{max} = 31.62 × 0.05 $$ $$ v_{max} = 1.58m/s $$ Also, the maximum acceleration attained at end point is given by: $$ a = \omega ² A $$ $$ a = 31.62² × 0.05 $$ $$ a = 49.99m/s² $$