We know that, Force F = m a = 5 × 0.4 = 2 N. The spring constant is given as: So the value of can be anything depending upon the position of the particle at t = 0. Just enter your phone number below. 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. It is a special case of oscillatory motion. ⁡. The body must experience a net Torque that is restoring in nature. The direction of this restoring force is always towards the mean position. In this type of oscillatory motion displacement, velocity and acceleration and force vary (w.r.t time) in a way that can be described by either sine (or) the cosine functions collectively called sinusoids. Angle made by the particle at t = 0 with the upper vertical axis is equal to φ (phase constant). ⇒ Variation of Kinetic Energy and Potential Energy in Simple Harmonic Motion with displacement: If a particle is moving with uniform speed along the circumference of a circle then the straight line motion of the foot of the perpendicular drawn from the particle on the diameter of the circle is called simple harmonic motion. Cloudflare Ray ID: 5f79dc465947d600 (a) Calculate t Now if we see the equation of position of the particle with respect to time, sin (ωt + Φ) – is the periodic function, whose period is T = 2π/ω, Which can be anything sine function or cosine function. Bonus: Use your free doubt session & get a free formula book! The point at which net force acting on the particle is zero. Hence the total energy of the particle in SHM is constant and it is independent of the instantaneous displacement. An example of such a scenario is simple pendulum. However, if the maximum displacement about the mean position is small, then the motion can be very often approximated to SHM. 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. If it is slightly pushed from its mean position and released, it makes angular oscillations. That is why it is called as initial phase of the particle. A pendulum beating seconds is taken from the earth to a planet having both the mass and radius half as compared to the earth. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. A 0.500 kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.5 cm. d2x→dt2=−ω2x→\frac{{{d}^{2}}\overrightarrow{x}}{d{{t}^{2}}}=-{{\omega }^{2}}\overrightarrow{x}dt2d2x​=−ω2x. Its time period of oscillation on the earth is $T.$ What will be its time period at the centre of the earth ? The restoring force or acceleration acting on the particle should always be proportional to the displacement of the particle and directed towards the equilibrium position.

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