Displacement, velocity, and acceleration in SHM




We still need to find the displacement x as a function of time for a harmonic oscillator. Equation (14.4) for a body in SHM along the x-axis is identical to Eq. (14.8) for the x-coordinate of the reference point in uniform circular motion with constant angular speed w = (k)1/2/m. Hence Eq. (14.5), x = Acos Ø, describes the x-coordinate for both situations. If at t = 0 the phasor OQ makes an angle f (the Greek letter phi) with the positive x-axis, then at any later time t this angle is Ø = wt + Ø. We substitute this into Eq. (14.5) to obtain

Displacement, velocity, and acceleration in SHM

 

 

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Figure 14.9 shows a graph of Eq. (14.13) for the particular case f = 0. We could also have written Eq. (14.13) in terms of a sine function rather than a cosine by using the identity cosa = sin1a + p/22. In simple harmonic motion the displacement is a periodic, sinusoidal function of time. There are many other periodic functions, but none so simple as a sine or cosine function.

Displacement, velocity, and acceleration in SHM

 

 

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The value of the cosine function is always between -1 and 1, so in Eq. (14.13), x is always between -A and A. This confirms that A is the amplitude of the motion.

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Displacement, velocity, and acceleration in SHM

 

 

 

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The cosine function in Eq. (14.13) repeats itself whenever time t increases by one period T, or when vt + f increases by 2p radians. Thus, if we start at time t = 0, the time T to complete one cycle is

Displacement, velocity, and acceleration in SHM

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which is just Eq. (14.12). Changing either m or k changes the period T (Figs. 14.10a and 14.10b), but T does not depend on the amplitude A (Fig. 14.10c).

The constant f in Eq. (14.13) is called the phase angle. It tells us at what point in the cycle the motion was at t = 0 (equivalent to where around the circle the point Q was at t = 0). We denote the displacement at t = 0 by x0. Putting t = 0 and x = x0 in Eq. (14.13), we get

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                     x0 = AcosØ                                  (14.14)

If f = 0, then x0 = Acos 0 = A, and the body starts at its maximum positive displacement. If f = p, then x0 = Acosp = -A, and the particle starts at its maximum negative displacement. If f = p>2, then x0 = Acos1p>22 = 0, and the particle is initially at the origin. Figure 14.11 shows the displacement x versus time for three different phase angles.

Displacement, velocity, and acceleration in SHM

 

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We find the velocity vx and acceleration ax as functions of time for a harmonic oscillator by taking derivatives of Eq. (14.13) with respect to time:

Displacement, velocity, and acceleration in SHM

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The velocity vx oscillates between vmax = +vA and -vmax = -vA, and the acceleration ax oscillates between amax = +v2 A and -amax = -v2 A (Fig. 14.12). Comparing Eq. (14.16) with Eq. (14.13) and recalling that v2 = k>m from Eq. (14.9), we see that

 

Displacement, velocity, and acceleration in SHM

 

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which is just Eq. (14.4) for simple harmonic motion. This confirms that Eq. (14.13) for x as a function of time is correct.

Displacement, velocity, and acceleration in SHM

 

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We actually derived Eq. (14.16) earlier in a geometrical way by taking the x-component of the acceleration vector of the reference point Q. This was done in Fig. 14.6b and Eq. (14.7) (recall that u = vt + f). In the same way, we could have derived Eq. (14.15) by taking the x-component of the velocity vector of Q, as shown in Fig. 14.6b. We’ll leave the details for you to work out.

Note that the sinusoidal graph of displacement versus time (Fig. 14.12a) is shifted by one-quarter period from the graph of velocity versus time (Fig. 14.12b) and by one-half period from the graph of acceleration versus time (Fig. 14.12c).

Displacement, velocity, and acceleration in SHM

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 14.13 shows why this is so. When the body is passing through the equilibrium position so that x = 0, the velocity equals either vmax or -vmax (depending on which way the body is moving) and the acceleration is zero. When the body is at either its most positive displacement, x = +A, or its most negative displacement, x = -A, the velocity is zero and the body is instantaneously at rest. At these points, the restoring force Fx = -kx and the acceleration of the body have their maximum magnitudes. At x = +A the acceleration is negative and equal to -amax. At x = -A the acceleration is positive: ax = +amax.

Here’s how we can determine the amplitude A and phase angle f for an oscillating body if we are given its initial displacement x0 and initial velocity v0x. The initial velocity v0x is the velocity at time t = 0; putting vx = v0x and t = 0 in Eq. (14.15), we find,

Displacement, velocity, and acceleration in SHM

 

 

To find f, we divide Eq. (14.17) by Eq. (14.14). This eliminates A and gives an equation that we can solve for f:

Displacement, velocity, and acceleration in SHM

 

 

 

 

It is also easy to find the amplitude A if we are given x0 and v0x. We’ll sketch the derivation, and you can fill in the details. Square Eq. (14.14); then divide Eq. (14.17) by v, square it, and add to the square of Eq. (14.14). The right side will be A2 1sin2f + cos2f2, which is equal to A2 . The final result is:

Displacement, velocity, and acceleration in SHM

 

 

Note that when the body has both an initial displacement x0 and a nonzero initial velocity v0x, the amplitude A is not equal to the initial displacement. That’s reasonable; if you start the body at a positive x0 but give it a positive velocity v0x, it will go farther than x0 before it turns and comes back, and so A 7 x0.



Frequently Asked Questions

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Ans: the period and frequency of simple harmonic motion are completely determined by the mass m and the force constant k. In simple harmonic motion the period and frequency do not depend on the amplitude A. view more..
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Ans: To explore the properties of simple harmonic motion, we must express the displacement x of the oscillating body as a function of time, x1t2. view more..
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Ans: The simplest kind of oscillation occurs when the restoring force Fx is directly proportional to the displacement from equilibrium x. This happens if the spring in Figs. 14.1 and 14.2 is an ideal one that obeys Hooke’s law view more..
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Ans: We still need to find the displacement x as a function of time for a harmonic oscillator. Equation (14.4) for a body in SHM along the x-axis is identical to Eq. (14.8) for the x-coordinate of the reference point in uniform circular motion with constant angular speed v = 2k/m view more..
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Ans: We can learn even more about simple harmonic motion by using energy considerations. The only horizontal force on the body in SHM in Figs. 14.2 and 14.13 is the conservative force exerted by an ideal spring. The vertical forces do no work, so the total mechanical energy of the system is conserved. We also assume that the mass of the spring itself is negligible. view more..
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Ans: the energy quantities E, K, and U at x = 0, x = ±A/2, and x = ±A. Figure 14.15 is a graphical display of Eq. (14.21); energy (kinetic, potential, and total) is plotted vertically and the coordinate x is plotted horizontally. The parabolic curve in Fig. 14.15a represents the potential energy U = 1/2 kx2 . The horizontal line represents the total mechanical energy E, which is constant and does not vary with x. view more..
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Ans: PROBLEM SOLVING STRATEGY ON ENERGY MOMENTUM OF SHM view more..
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Ans: So far, we’ve looked at a grand total of one situation in which simple harmonic motion (SHM) occurs: a body attached to an ideal horizontal spring. But SHM can occur in any system in which there is a restoring force that is directly proportional to the displacement from equilibrium, as given by Eq. (14.3), Fx = -kx view more..
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Ans: A mechanical watch keeps time based on the oscillations of a balance wheel (Fig. 14.19). The wheel has a moment of inertia I about its axis. A coil spring exerts a restoring torque tz that is proportional to the angular displacement u from the equilibrium position. We write tz = -ku, where k (the Greek letter kappa) is a constant called the torsion constant. Using the rotational analog of Newton’s second law for a rigid body, gtz = Iaz = I d2 u>dt2 view more..
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Ans: The following discussion of the vibrations of molecules uses the binomial theorem. If you aren’t familiar with this theorem, you should read about it in the appropriate section of a math textbook. view more..
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Ans: A simple pendulum is an idealized model consisting of a point mass suspended by a massless, unstretchable string. When the point mass is pulled to one side of its straight-down equilibrium position and released, it oscillates about the equilibrium position. view more..
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Ans: A physical pendulum is any real pendulum that uses an extended body, as contrasted to the idealized simple pendulum with all of its mass concentrated at a point. F view more..




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