सरल आवर्त गति | Simple Harmonic Motion | NEET 2021 | Physics | Crash Course | Doubtnut | Vipin Sir

Time Period Of Simple Harmonic Motion || Target JEE Main / NEET 2019

9.30 PM JEE Mains 2020 - Physics | Simple Harmonic Motion - 30 Days Score Booster Course

General physics || Superposition OF Simple Harmonic Motion

JEE Mains 2020 - 30 Days Revision | Simple Harmonic Motion - Most Important Questions | Physics

In physical pendulum, the time period for small oscillation is given by T=2pisqrt((I)/(Mgd)) where I is the moment of inertial of the body about an axis passing through a pivoted point O and perpendicular to the plane of oscillation and d is the separation point between centre of gravity and the pivoted point. In the physical pendulum a speacial point exists where if we concentrate the entire mass of body, the resulting simple pendulum (w.r.t. pivot point O) will have the same time period as that of physical pendulum This point is termed centre of oscillation. T=2pisqrt((I)/(Mgd))=2pisqrt((L)/(g)) Moreover, this point possesses two other important remarkable properties: Property I: Time period of physical pendulum about the centre of oscillation (if it would be pivoted) is same as in the original case. Property II: If an impulse is applied at the centre of oscillatioin in the plane of oscillation, the effect of this impulse at pivoted point is zero. Because of this property, this point is also known as the centre of percussion. From the given information answer the following question: Q. A uniform rod of mass M and length L is pivoted about point O as shown in Figgt It is slightly rotated from its mean position so that it performs angular simple harmonic motion. For this physical pendulum, determine the time period oscillation.

Simple Harmonic Motion(सरल आवर्त गति)|Energy Of SHM(सरल आवर्त गति में ऊर्जा )|Questions(प्रश्न )|Summary

Simple Harmonic Motion(सरल आवर्त गति)|Velocity Of SHM(एसएचएम का वेग)|Acceleration In SHM(एसएचएम में त्वरण)|Questions(प्रश्न )|OMR|Summary