At point A v = 0 [x = A] the equation (1) becomes, O = −ω2A22+c\frac{-{{\omega }^{2}}{{A}^{2}}}{2}+c2−ω2A2​+c, c = ω2A22\frac{{{\omega }^{2}}{{A}^{2}}}{2}2ω2A2​, ⇒ v2=−ω2x2+ω2A2{{v}^{2}}=-{{\omega }^{2}}{{x}^{2}}+{{\omega }^{2}}{{A}^{2}}v2=−ω2x2+ω2A2, ⇒ v2=ω2(A2−x2){{v}^{2}}={{\omega }^{2}}\left( {{A}^{2}}-{{x}^{2}} \right)v2=ω2(A2−x2), v = ω2(A2−x2)\sqrt{{{\omega }^{2}}\left( {{A}^{2}}-{{x}^{2}} \right)}ω2(A2−x2)​, v = ωA2−x2\omega \sqrt{{{A}^{2}}-{{x}^{2}}}ωA2−x2​ … (2), where, v is the velocity of the particle executing simple harmonic motion from definition instantaneous velocity, v = dxdt=ωA2−x2\frac{dx}{dt}=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}dtdx​=ωA2−x2​, ⇒ ∫dxA2−x2=∫0tωdt\int{\frac{dx}{\sqrt{{{A}^{2}}-{{x}^{2}}}}}=\int\limits_{0}^{t}{\omega dt}∫A2−x2​dx​=0∫t​ωdt, ⇒ sin⁡−1(xA)=ωt+ϕ{{\sin }^{-1}}\left( \frac{x}{A} \right)=\omega t+\phisin−1(Ax​)=ωt+ϕ. So the value of can be anything depending upon the position of the particle at t = 0. The equilibrium position is shown as a black dot and is the point where the force is equal to zero. This is usually not exactly valid for many systems under simple harmonic oscillations. Its amplitude is $A.$ The potential energy of the particle is equal to the kinetic energy, when the displacement of the particle from the mean position is, The potential energy of a particle executing SHM varies sinusoidally. Frequency = 1/T and, angular frequency ω = 2πf = 2π/T. From the expression of particle position as a function of time: We can find particles, displacement (x→),\left( \overrightarrow{x} \right), (x),velocity (v→)\left( \overrightarrow{v} \right)(v) and acceleration as follows. The object will keep on moving between two extreme points about a fixed point is called mean position (or) equilibrium position along any path. To derive an equation for the period and the frequency, we must first define and analyze the equations of motion. The phases of the two SHM differ by π/2. It is a special case of oscillatory motion. The expression, position of a particle as a function of time. SHM or Simple Harmonic Motion can be classified into two types. Similarly, the foot of the perpendicular on the y-axis is called vertical phasor. When ω = 1 then, the curve between v and x will be circular. Thus, we see that the uniform circular motion is the combination of two mutually perpendicular linear harmonic oscillation. However, if the maximum displacement about the mean position is small, then the motion can be very often approximated to SHM. The equation (3) – equation of position of a particle as a function of time. The particle is at position P at t = 0 and revolves with a constant angular velocity (ω) along a circle. . Equation of SHM is given by F =−kx. Hence the total energy of the particle in SHM is constant and it is independent of the instantaneous displacement. This is the differential equation of an angular Simple Harmonic Motion. To and fro motion of a particle about a mean position is called an oscillatory motion in which a particle moves on either side of equilibrium (or) mean position is an oscillatory motion. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. i.e.sin⁡−1(x0A)=ϕ{{\sin }^{-1}}\left( \frac{{{x}_{0}}}{A} \right)=\phisin−1(Ax0​​)=ϕ initial phase of the particle, Case 3: If the particle is at one of its extreme position x = A at t = 0, ⇒ sin⁡−1(AA)=ϕ{{\sin }^{-1}}\left( \frac{A}{A} \right)=\phisin−1(AA​)=ϕ, ⇒ sin⁡−1(1)=ϕ{{\sin }^{-1}}\left( 1 \right)=\phisin−1(1)=ϕ. The total work done by the restoring force in displacing the particle from (x = 0) (mean position) to x = x: When the particle has been displaced from x to x + dx the work done by restoring force is, w = ∫dw=∫0x−kxdx=−kx22\int{dw}=\int\limits_{0}^{x}{-kxdx=\frac{-k{{x}^{2}}}{2}}∫dw=0∫x​−kxdx=2−kx2​, = −mω2x22-\frac{m{{\omega }^{2}}{{x}^{2}}}{2}−2mω2x2​ [ k=mω2]\left[ \,k=m{{\omega }^{2}} \right][k=mω2], = −mω22A2sin⁡2(ωt+ϕ)-\frac{m{{\omega }^{2}}}{2}{{A}^{2}}{{\sin }^{2}}\left( \omega t+\phi \right)−2mω2​A2sin2(ωt+ϕ), Potential Energy = -(work done by restoring force), Potential Energy = mω2x22=mω2A22sin⁡2(ωt+ϕ)\frac{m{{\omega }^{2}}{{x}^{2}}}{2}=\frac{m{{\omega }^{2}}{{A}^{2}}}{2}{{\sin }^{2}}\left( \omega t+\phi \right)2mω2x2​=2mω2A2​sin2(ωt+ϕ), E = 12mω2(A2−x2)+12mω2x2\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{x}^{2}} \right)+\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}21​mω2(A2−x2)+21​mω2x2, E = 12mω2A2\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}21​mω2A2. d2x/dt2 + ω2x = 0, which is the differential equation for linear simple harmonic motion. The horizontal component of the velocity of a particle gives you the velocity of a particle performing the simple harmonic motion. Determine its spring constant. Solution of this equation is angular position of the particle with respect to time. Any oscillatory motion which is not simple Harmonic can be expressed as a superposition of several harmonic motions of different frequencies. Simple Harmonic Motion or SHM is defined as a motion in which the restoring force is directly proportional to the displacement of the body from its mean position.


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