Topic 4.1 Extended A – Mass-spring system SHM

1. Topic 4.1 ExtendedA – Mass-spring system SHM I'm sure you don't want to cover all the material for simple harmonic motion again, but it is time to look again at physics from the Newtonian angle (which is the calculus angle). In fact, from the calculus angle, the whole "equations of motion" thing is much simpler and more elegant.

2. FYI: You will also see the "dot" notation for the time derivative. Thus x'x, and x"  x. . : Topic 4.1 ExtendedA – Mass-spring system SHM TWO INTERESTING FUNCTIONS: Consider the function x(t) = Asin(ωt+), where A, ω, and  are constants. If we take the first and second derivatives of x we get, respectively x (t) = Acos(ωt+) x (t) = Asin(ωt+) x’(t) = -ωAsin(ωt+) x’(t) = ωAcos(ωt+) x”(t) = -ω2Acos(ωt+) x”(t) = -ω2Asin(ωt+) x”(t) = -ω2x(t) x”(t) = -ω2x(t) FYI: The second derivative reproduces the original function. FYI: The second derivative of x(t) = Acos(ωt+)also reproduces the original function.

3. The solutions to the SHM equation F = -kx FYI: A differential equation is an equation that has derivatives in it. Topic 4.1 ExtendedA – Mass-spring system SHM FYI: We didn't need to look at shadows of uniform circular motion to find the equations of motion. SOLVING THE DIFFERENTIAL EQUATION F = -kx: Consider the following analysis… F = -kx ma = -kx Why can we write this? mx” = -kx k m x” = - x k m Provided ω2= x” = -ω2x The above is a differential equation whose solutions we found in the beginning of this section. Thus F = -kx has the solution x(t) = Asin(ωt+) k m Provided ω= or x(t) = Acos(ωt+)

4. x t Topic 4.1 ExtendedA – Mass-spring system SHM SOME PROPERTIES OF SINUSOIDAL FUNCTIONS: To illustrate the effect of A, consider the two displacements:x = 1sin(ωt) and x = 2sin(ωt): x = 2sin(ωt) x = 1sin(ωt) The amplitude is affected by varying A. The frequency does not change.

5. Tblue Tblue Tblue x Tred Tred t Topic 4.1 ExtendedA – Mass-spring system SHM SOME PROPERTIES OF SINUSOIDAL FUNCTIONS: To illustrate the effect of ω, consider the two displacements:x = sin(ωt) and x = sin(2ωt): x = sin(2ωt) x = sin(ωt) The frequency and period are affected by varying ω, (not the amplitude). If the frequency is doubled, the period is halved:

6. x t Π 2 x = sin(ωt +) Π 2 sin(ω·0+ ) = 1. Topic 4.1 ExtendedA – Mass-spring system SHM SOME PROPERTIES OF SINUSOIDAL FUNCTIONS: To illustrate the effect of , consider the two displacements:x = sin(ωt+0) and x = sin(ωt+): Π 2 x = sin(ωt) The phase constant affects only the starting point of the graph, and nothing else. The effect of adding the phase constant is to shift the graph LEFT. For example, sin(ω·0+0)= 0 , whereas

7. k k m Topic 4.1 ExtendedA – Mass-spring system SHM EXAMPLES USING THIS TECHNIQUE: Suppose a block of mass m is connected to two springs, each having a spring constant of k. Find the angular frequency ω of the system. Since displacing the mass a distance x will cause opposing forces from two springs of –kx eachwe have ma = -2kx mx” = -2kx 2k m x” = - x 2k m x” = -ω2x whereω= FYI: Now you can find T and f if you want, and use the general solutions for the differential equation x” = -ω2x.

8. k k m FYI: The critical difference between this problem and the previous one is that only ONE spring is in contact with the mass. Topic 4.1 ExtendedA – Mass-spring system SHM EXAMPLES USING THIS TECHNIQUE: Suppose a block of mass m is connected to two springs, each having a spring constant of k. Find the angular frequency ω of the system. Since displacing the mass a distance x will cause an opposing force from one spring of –kx/2 we have ma = -kx/2 mx” = -kx/2 k 2m x” = - x k 2m x” = -ω2x whereω= FYI: Now you can find T and f if you want, and use the general solutions for the differential equation x” = -ω2x.

9. Topic 4.1 ExtendedA – Mass-spring system SHM THE BIG THREE: If x(t) = Asin(ωt+), or x(t) = Acos(ωt+), we have x (t) = Acos(ωt+) x (t) = Asin(ωt+) v (t) x’(t) = -ωAsin(ωt+) x’(t) = ωAcos(ωt+) v (t) a (t) x”(t) = -ω2Acos(ωt+) a (t) x”(t) = -ω2Asin(ωt+) FYI: Recall that x' = v. FYI: Recall that x" = v' = a. Once you have the position function, you can find v and a simply by taking derivatives. You don't have to memorize a bunch of equations.

10. Topic 4.1 ExtendedA – Mass-spring system SHM A 2-kg mass has a position function of x(t) = 0.5 cos(5t + /4), where x is in meters and t is in seconds. (a) Find v(t) and a(t): dx dt Since x = 0.5cos(5t + /4) and since v = , all we have to do is take the derivative of x: x = 0.5cos(5t + /4) dx dt = 0.5[-sin(5t + /4)][5] Why? v = -2.5sin(5t + /4) dv dt Since v = -2.5sin(5t + /4) and since a = , all we have to do is take the derivative of v: dv dt = -2.5[cos(5t + /4)][5] Why? a = -12.5cos(5t + /4)

11. k k  = 5 = m 3 k 25 = 3 Topic 4.1 ExtendedA – Mass-spring system SHM A 3-kg mass has a position function given by x(t) = 0.5 cos(5t + /4), where x is in meters and t is in seconds. (b) Find the spring constant: we have Since  = 5 and since k = 75 n/m

12. Topic 4.1 ExtendedA – Mass-spring system SHM A 3-kg mass has a position function given by x(t) = 0.5 cos(5t + /4), where x is in meters and t is in seconds. (c) Find the "big three" at t = 0 s: x = 0.5cos(5·0 + /4) = 0.35 m v = -2.5sin(5·0 + /4) = -1.77 m/s a = -12.5cos(5·0 + /4) = -6.94 m/s2