\end{array}\], Thus the position as a function of time can be written as, \[x(t)=A \cos \left(\omega_{0} t+\phi\right)\], In Equation (23.2.25) the quantity \(\omega_{0} t+\phi\) is called the phase, and \(\phi\) is called the phase constant. one end of the spring be attached to a wall and let the object move horizontally Assume the spring is stretched a distance A from its equilibrium position and then released. displacement. CONTACT An object is undergoing simple harmonic motion (SHM) if; the acceleration of the object is directly proportional to its displacement from its equilibrium position. Although we had previously specified \(x_{0}>0 \text { and } v_{x, 0}>0\), Equation (23.2.21) is seen to be a valid solution of the SHO equation for any values of \(x_{0}\) and \(v_{x, 0}\). displaced upward by a distance x, then the total force on the mass is mg - k(x0 (b)  the maximum speed of the mass, and a(t) = -ω2Acos(ωt + φ) = -ω2x. Let $v$ be the speed of the wave. Whenever you encounter a It obeys Hooke's law, F = -kx, with k = mω2. \frac{d^{2}}{d t^{2}} x(t) &=-\frac{k}{m} x_{1}(t)+-\frac{k}{m} x_{2}(t)=-\frac{k}{m}\left(x_{1}(t)+x_{2}(t)\right) \\ per unit time. With what amplitude does the particle oscillate? Equation of Simple Harmonic Motion AMPLITUDE. MECHANICS then the motion of the object is simple harmonic motion. Because \(\cos (0)=1 \text { and } \sin (0)=0\), the initial position at time t = 0 is, The x -component of the velocity at time t = 0 is, \[v_{x, 0}=v_{x}(t=0)=-\omega_{0} C \sin (0)+\omega_{0} D \cos (0)=\omega_{0} D\], \[C=x_{0} \text { and } D=\frac{v_{x, 0}}{\omega_{0}}\], and the x -component of the velocity of the object-spring system is, \[v_{x}(t)=-\sqrt{\frac{k}{m}} x_{0} \sin (\sqrt{\frac{k}{m}} t)+v_{x, 0} \cos (\sqrt{\frac{k}{m}} t)\]. Equation (23.2.2) can be solved from energy considerations or other advanced techniques but instead we shall first guess the solution and then verify that the guess satisfies the SHO differential equation (see Appendix 22.3.A for a derivation of the solution). Note that the magnitude of velocity $v_\text{y}$ is not the wave speed. First you know the wave equation for the wave travelling in positive x-direction from Eq. frequency of the motion is, If the only force acting on an object with mass m is a Hooke's law force, with ω = b. This occurs when \(\omega_{0} t+\phi=2 \pi n\) where \(n=0,\pm 1,\pm 2, \cdots\) The maximum value associated with n = 0 occurs when \(\omega_{0} t+\phi=0 \text { or } t=-\phi / \omega_{0}\). + x) = -kx directed towards the equilibrium position. This can be verified by multiplying the equation by , and then making use of the fact that . the vertical position and the horizontal position. a force on the object. In this case, the constant of proportionality is k/m. When \(\phi<0\) the, function \(x(t)=A \cos \left(\omega_{0} t+\phi\right)\) reaches its maximum value at a later time \(t=T / 8\) than the function \(x(t)=A \cos \left(\omega_{0} t\right)\) as shown in Figure 23.4c. \text{or,}\quad \frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}&=\frac{1}{{{v}^{2}}}\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}} \tag{11} \label{11} Periodic function f (t) with period T are those functions of the variable ‘t’ … The simple harmonic oscillator equation, , is a linear differential equation, which means that if is a solution then so is , where is an arbitrary constant. Suppose mass of a particle executing simple harmonic motion is ‘m’ and if at any moment its displacement and acceleration are respectively x and a, then according to definition, \frac{d^{2}}{d t^{2}} x_{1}(t)=-\frac{k}{m} x_{1}(t) \\ \eqref{11} is called linear wave equation which gives total description of wave motion. The force is. Let Path length or range is twice the amplitude and in one oscillation body covers a total distance of ‘4a’. \frac{{{\partial }^{2}}y/\partial {{t}^{2}}}{{{\partial }^{2}}y/d{{x}^{2}}}&=\frac{{{\omega }^{2}}}{{{k}^{2}}} \\ Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Equation (3) gives the displacement of the... TIME PERIOD. Now you can replace $t$ in $y=f(x)=A\cos \omega t$ by $t - x/v$ and, \[y=A\cos \omega \left( t-\frac{x}{v} \right)=A\cos \omega \left( \frac{x}{v}-t \right) \tag{1} \label{1}\], We know that $\omega = 2 \pi f$, $f = 1/T$ and $T = \lambda /v$, and you can simplify the above equation in the form, \[y=A\cos 2\pi \left( \frac{x}{\lambda } -\frac{t}{T} \right) \tag{2} \label{2}\], We define a quantity called wave number denoted by $k$ which is $k = 2\pi / \lambda$ and you know that $2\pi/T$ is the angular frequency $\omega$ and therefore we can again write Eq. 23.2.1 General Solution of Simple Harmonic Oscillator Equation; Example 23.1: Phase and Amplitude ; Example 23.2: Block-Spring System; Our first example of a system that demonstrates simple harmonic motion is a springobject system on a frictionless surface, shown in Figure 23.2. \tan \phi=\frac{\sin \phi}{\cos \phi}=\frac{-D / A}{C / A}=-\frac{D}{C} m/s. The motion repeats. + φ) + cos2(ωt + φ)) = ½mω2A2. Equation (23.2.1) is a second order linear differential equation, in which the second derivative of the dependent variable is proportional to the negative of the dependent variable, \[\frac{d^{2} x}{d t^{2}}=-\frac{k}{m} x\]. The velocity is zero at maximum displacement, and Thus the linear combination \(x(t)=x_{1}(t)+x_{2}(t)\) is also a solution of the SHO equation, Equation (23.2.1). Missed the LibreFest? This equation is obtained for a special case of wave called simple harmonic wave but it is equally true for other periodic or non-periodic waves. rad/s. Watch the recordings here on Youtube! proportional to the displacement, but in the opposite direction. For example, and in Equation , or and in Equation . conservative force. The same equations have the same solutions. oscillates back and forth. differential equation of the form d2x/dt2 = -b2x, you know This occurs at time \(t_{1}\) satisfying, \[\sqrt{\frac{k}{m}} t_{1}=\frac{\pi}{2}, \quad t_{1}=\frac{\pi}{2} \sqrt{\frac{m}{k}}=\frac{T}{4}\], The x -component of the velocity at time \(t_{1}\) is then, \[v_{x}\left(t_{1}\right)=-\sqrt{\frac{k}{m}} x_{0} \sin \left(\sqrt{\frac{k}{m}} t_{1}\right)=-\sqrt{\frac{k}{m}} x_{0} \sin (\pi / 2)=-\sqrt{\frac{k}{m}} x_{0}=-\omega_{0} x_{0}\]. The angular frequency ω = (k/m)½ is the same The angular frequency is measured in radians per second. oscillations per second and an amplitude of 5 cm. The graph of displacement $y$ of particles with position on the x-axis at an instant of time $t$ is shown in Figure 1. \eqref{11} is called linear wave equation which gives total description of wave motion. Solution: A 20 g particle moves in simple harmonic motion with a frequency of 3 acceleration is in the direction of its velocity. - x) = kx, directed towards the equilibrium position. Where does that occur? Note that at \(t=0\), the position of the object is \(x_{0} \equiv x(t=0)=A\) since \(\cos (0)=1\) and the velocity is \(v_{x, 0} \equiv v_{x}(t=0)=0\) since \(\sin (0)=0\). The spring is initially stretched a distance \(l_{0}\) and given some initial speed \(v_{0}\) to the right away from the equilibrium position. \eqref{6}, you can get a form: \[\begin{align*} \eqref{9} with respect to $x$ keeping $t$ constant, you'll get the curvature of the curve at position $x$: \[\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}=-A{{k}^{2}}\cos (kx-\omega t) \tag{10} \label{10}\].