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
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.
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.
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
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
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.
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:
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
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.
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).
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,
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:
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:
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
- Nature of physics
- Solving Physics Problems
- Standards and Units
- Using and Converting Units
- Uncertainty and significant figures
- Estimates and order of magnitudes
- Vectors and vector addition
- Equilibrium and Elasticity
- Conditions for equilibrium
- Center of gravity
- finding and using the Center of gravity
- solving rigid-body equilibrium problems
- SOLVED EXAMPLES ON EQUILIBRIUM
- stress, strain, and elastic moduLi
- tensile and Compressive stress and strain