Simple harmonic motion And its application
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This presentation introduces simple harmonic motion (SHM) and its applications. SHM describes a back-and-forth motion where the restoring force is directly proportional to displacement from the equilibrium position. One example is the motion of a spring according to Hooke's law, where the force is proportional to displacement. SHM has a differential equation and solutions that describe oscillations with a characteristic frequency and period. Common applications that exhibit SHM include springs, pendulums, and torsional oscillators.
Simple harmonic motion And its application
- 2. SIMPLE HARMONIC MOTION AND ITS APPLICATION PRESENTED BY MD. SUMON SARDER ID: 161-15-953
- 3. SIMPLE HARMONIC MOTION BACK AND FORTH MOTION WHICH IS CAUSED BY A FORCE THAT IS DIRECTLY PROPORTIONAL TO THE DISPLACEMENT. THE DISPLACEMENT CENTERS AROUND AN EQUILIBRIUM POSITION. xFs
- 4. Springs – Hooke’s Law kxorkxF k k xF s s N/m):nitConstant(USpring alityProportionofConstant One of the simplest type of simple harmonic motion is called Hooke's Law. This is primarily in reference to SPRINGS.
- 5. Simple Harmonic Motion (SHM) Hence, In SHM the restoring force is proportional to the displacement and acts in the opposite direction of that displacement
- 6. Differential Equation of SHM
- 7. Value of 𝜔 Putting the solution in the differential equation we get the value of 𝜔 𝑑2 𝑥 𝑑𝑡2 + 𝑘 𝑚 𝑥 = 0 −𝜔2 𝐴𝑐𝑜𝑠 𝜔𝑡 + 𝜙 + 𝑘 𝑚 𝐴𝑐𝑜𝑠 𝜔𝑡 + 𝜙 = 0 𝜔 = 𝑘 𝑚
- 8. Time Period of SHM If we increase the time t in the solution of SHM by 2𝜋/𝜔, the function becomes, 𝑥 = 𝐴𝑐𝑜𝑠 𝜔 𝑡 + 2𝜋 𝜔 + 𝜙 = 𝐴𝑐𝑜𝑠 𝜔𝑡 + 2𝜋 + 𝜙 = 𝐴𝑐𝑜𝑠 𝜔𝑡 + 𝜙 i.e. the function repeats itself after a time 2𝜋/𝜔 Therefore 2𝜋/𝜔 is the period of the motion, T. 𝑇 = 2𝜋 𝜔 𝜔 = 2𝜋 𝑇 = 2𝜋𝑓, is the angular frequency.
- 11. Total Energy in SHM
- 12. Applications of SHM As per Hooke’s law the restoring force on a spring is F=-kx, the exact condition for simple harmonic oscillation.
- 13. Simple Pendulum
- 15. The Physical Pendulum If we locate the pivot far from the object , using a weightless cord of length L, we would have, 𝐼 = 𝑀𝐿2 𝑎𝑛𝑑 ℎ = 𝐿 Then, 𝑇 = 2𝜋 𝐼 𝑀𝑔ℎ = 2𝜋 𝑀𝐿2 𝑀𝑔𝐿 = 2𝜋 𝐿 𝑔 Which is the period of a simple pendulum!