# We divide by m and let ω pk m to rewrite the

• shamessaimun
• 322

This preview shows page 106 - 108 out of 322 pages.

We divide by m and let ω 0 p k / m to rewrite the equation as x 00 + ω 2 0 x 0 . The general solution to this equation is x ( t ) A cos ( ω 0 t ) + B sin ( ω 0 t ) . By a trigonometric identity A cos ( ω 0 t ) + B sin ( ω 0 t ) C cos ( ω 0 t - γ ) , for two different constants C and γ . It is not hard to compute that C A 2 + B 2 and tan γ B / A . Therefore, we let C and γ be our arbitrary constants and write x ( t ) C cos ( ω 0 t - γ ) . Exercise 2.4.1 : Justify the identity A cos ( ω 0 t ) + B sin ( ω 0 t ) C cos ( ω 0 t - γ ) and verify the equations for C and γ . Hint Start with cos ( α - β ) cos ( α ) cos ( β ) + sin ( α ) sin ( β ) and multiply by C . Then what should α and β be While it is generally easier to use the first form with A and B to solve for the initial conditions, the second form is much more natural. The constants C and γ have nice physical interpretation. Write the solution as x ( t ) C cos ( ω 0 t - γ ) . This is a pure-frequency oscillation (a sine wave). The amplitude is C , ω 0 is the (angular) frequency , and γ is the so-called phase shift . The phase shift just shifts the graph left or right. We call ω 0 the natural (angular) frequency . This entire setup is called simple harmonic motion . Let us pause to explain the word angular before the word frequency . The units of ω 0 are radians per unit time, not cycles per unit time as is the usual measure of frequency. Because one cycle is 2 π radians, the usual frequency is given by ω 0 2 π . It is simply a matter of where we put the constant 2 π , and that is a matter of taste. The period of the motion is one over the frequency (in cycles per unit time) and hence 2 π ω 0 . That is the amount of time it takes to complete one full cycle. Example 2.4.1: Suppose that m 2 kg and k 8 N / m . The whole mass and spring setup is sitting on a truck that was traveling at 1 m / s . The truck crashes and hence stops. The mass was held in place 0.5 meters forward from the rest position. During the crash the mass gets loose. That is, the mass is now moving forward at 1 m / s , while the other end of the spring is held in place. The mass therefore starts oscillating. What is the
. . MECHANICAL VIBRATIONS 107 frequency of the resulting oscillation? What is the amplitude? The units are the mks units (meters-kilograms-seconds). The setup means that the mass was at half a meter in the positive direction during the crash and relative to the wall the spring is mounted to, the mass was moving forward (in the positive direction) at 1 m / s . This gives us the initial conditions. So the equation with initial conditions is 2 x 00 + 8 x 0 , x ( 0 ) 0 . 5 , x 0 ( 0 ) 1 . We directly compute ω 0 p k / m 4 2 . Hence the angular frequency is 2. The usual frequency in Hertz (cycles per second) is 2 / 2 π 1 / π 0 . 318 . The general solution is x ( t ) A cos ( 2 t ) + B sin ( 2 t ) . Letting x ( 0 ) 0 . 5 means A 0 . 5 . Then x 0 ( t ) - 2 ( 0 . 5 ) sin ( 2 t ) + 2 B cos ( 2 t ) . Letting x 0 ( 0 ) 1 we get B 0 . 5 . Therefore, the amplitude is C A 2 + B 2 0 . 25 + 0 . 25 0 . 5 0 . 707 . The solution is x ( t ) 0 . 5 cos ( 2 t ) + 0 . 5 sin ( 2 t ) .
• • • 