Coupon Accepted Successfully!


Periodic Motion: One Oscillator

Let's think about a mass m connected to a horizontal spring (resting length l0, spring constant k) which is connected to a wall. The mass is sitting on a frictionless floor (so we can ignore vertical forces). We stretch the spring to length l0 + A and let go. What happens?

At first the spring pulls the mass back towards equilibrium. As the mass goes faster, the displacement x and force Fspring decrease. When the spring is length l0, the mass is moving velocity –vmax. Because the length is l0 (and thus x = 0), the restoring force and acceleration are zero. The maximum displacement occurs when v = 0 and the magnitude of the spring force is very large. Figure 11-5 shows the movement at five times. Spend some time making sure you understand every entry on the table. In the first line the negative sign of the acceleration indicates the acceleration vector points left, the same as the force of the spring. This motion is called simple harmonic motion.

..\art 11 jpg\figure 11-td.jpg

Figure 11-5


Energy changes from potential to kinetic and back again, but the total energy is conserved. We have
The maximum displacement of the mass from equilibrium is the amplitude A. The period T is the time it takes for the system to go through one cycle, and it has units [s]. The frequency f is the number of cycles a system goes through in a unit of time, and it has units [1/s = Hertz = Hz]. We have
The frequency is related to the spring constant and the mass as follows:

Let's see if this equation makes sense. We would guess that a system with a stiff spring would have a high frequency (think about it), so it makes sense that k should be in the numerator. As k increases, f increases. Also a larger mass will decrease the frequency, so it makes sense that m is in the denominator. The square root is needed to make the units agree.

Let's consider a similar set up. In a pendulum we have a bob connected to a string or a light rod, which is connected to a ceiling of some sort. In this case, the restoring force is provided not by a spring but by a component of gravity. In Figure 11-6, the solid arrows are the two forces on the bob, and the dashed arrows show the gravitational force divided into components, along the supporting rod and perpendicular to it. The restoring force is the latter
F = mg sinθ  mgθ ...(7)
where sinθ is approximately equal to (measured in radians) for small angles. Note the similarity of this equation to equation (1). In both, the force is proportional to the displacement, so the motion is similar.
The frequency of a pendulum is given by

Note that the mass of the bob m does not appear in this equation, so a 10-kg mass swings on a 3-m string with the same period as a 0.1-kg mass swinging on a 3-m string. You need not memorize this equation, but you should be familiar with the fact that m does not appear in it.

..\art 11 jpg\figure 1-t.jpg

Figure 11-6

Test Your Skills Now!
Take a Quiz now
Reviewer Name