# Some system Executing Simple Harmonic Motion

A point mass suspended from a massless spring or placed on a frictionless horizontal plane attached with spring(fig) constitutes a linear harmonic spring pendulum.

If the point mass is pulled on one side and is released,it then executes a to and fro motion about the mean position.let x=0, indicate the position of the centre of the point mass when the spring is in equilibrium.the positions marked as –A and +A  indicate the maximum displacements to the left and the right of the mean position.

At any time t,if the displacement of the point mass from its mean position is x,the restoring force F acting on the block is,

F(x) = - kx

Where k=spring constant

Time period T = 2 \pi \sqrt{\frac{m}{k}}

frequency f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}

### Simple Pendulum View the Topic in this video From 0:39 To 12:40

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1. F(t) = ma = − mω2 x(t)
Where, k = mω2 is force constant.

2. At any displacement x,
Potential energy, U = \frac{1}{2}m\omega^{2}x^{2} = \frac{1}{2}kx^{2}
\tt =\frac{1}{2}kA^{2} \cos^{2}(\omega t + \phi)

3. Kinetic energy,
\tt K =\frac{1}{2}mv^{2} =\frac{1}{2} m\omega^{2} A^{2} \sin^{2}(\omega t + \phi)
\tt =\frac{1}{2}kA^{2} \sin^{2}(\omega t + \phi)

4. Total energy, E = K + U = \frac{1}{2}kA^{2}, It always remains constant even though K and U change with time

5. A block of mass m oscillating under the influence of restoring force F = −kx exhibits simple hormonic motion with angular frequency \omega = \sqrt{k/m} and time period T = 2 \pi \sqrt{\frac{m}{k}}

6. The motion of a simple pendulum is simple harmonic whose time period and frequency are given by
T = 2 \pi \sqrt{\frac{L}{g}} \ {\tt and} \ \nu = \frac{1}{2 \pi}\sqrt{\frac{g}{L}}