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S E C T I O N 15 . 2 • Mathematical Representation of Simple Harmonic Motion 461 same for all four springs. We see that the effective spring Hence, the frequency of vibration is, from Equation 15.14, To finalize the problem, note that the mass we used here is 1.18 Hz f ! 1 2)
√ k eff m ! 1 2)
√ 80 000 N/m 1 460 kg ! k
eff ! !
k ! 4 , 20 000 N/m ! 80 000 N/m end goes up when the back end goes down, the frequency What If? Suppose the two people exit the car on the side of the road. One of them pushes downward on the car and Answer The suspension system of the car is the same, but As we predicted conceptually, the frequency is a bit higher. f ! 1 2)
√ k eff m ! 1 2)
√ 80 000 N/m 1 300 kg ! 1.25 Hz Example 15.3 A Block–Spring System A 200-g block connected to a light spring for which the (A) Find the period of its motion. Solution From Equations 15.9 and 15.10, we know that the and the period is (B) Determine the maximum speed of the block. Solution We use Equation 15.17: (C) What is the maximum acceleration of the block? Solution We use Equation 15.18: (D) Express the position, speed, and acceleration as func- tions of time. Solution We find the phase constant from the initial condi- which tells us that ( ! 0. Thus, our solution is x ! A cos 't. " (1.25 m/s 2 )cos 5.00t a !"' 2 A cos 't ! " (0.250 m/s)sin 5.00t v !'A sin 't ! (0.050 0 m)cos 5.00t x !A cos 't ! x(0) ! A cos ( ! A 1.25 m/s 2 a
max ! ' 2 A ! (5.00 rad/s) 2 (5.00 , 10 " 2 m) ! 0.250 m/s v max ! ' A ! (5.00 rad/s)(5.00 , 10 " 2 m) ! 1.26 s - ! 2) ' ! 2) 5.00 rad/s ! ' ! √ k m ! √ 5.00 N/m 200 , 10 " 3 kg ! 5.00 rad/s What If? What if the block is released from the same initial position, x i ! 5.00 cm, but with an initial velocity of v i ! " 0.100 m/s? Which parts of the solution change and what are the new answers for those that do change? Answers Part (A) does not change—the period is indepen- Dividing Equation (2) by Equation (1) gives us the phase Now, Equation (1) allows us to find A: The new maximum speed is The new magnitude of the maximum acceleration is The new expressions for position, velocity, and acceleration As we saw in Chapters 7 and 8, many problems are easier to a ! "(1.35 m/s 2 )cos(5.00t % 0.12)) v ! "(0.269 m/s)sin(5.00t % 0.12)) x ! (0.053 9 m)cos(5.00t % 0.12)) a max ! ' 2 A ! (5.00 rad/s) 2 (5.39 , 10 " 2 m) ! 1.35 m/s 2 v max ! ' A ! (5.00 rad/s)(5.39 , 10 " 2 m) ! 0.269 m/s A ! x i cos
( ! 0.050 0 m cos(0.12)) ! 0.053 9 m ( ! 0.12) tan
( ! " v i ' x i ! " " 0.100 m (5.00 rad/s)(0.050 0 m) ! 0.400
" ' A sin ( A cos ( ! v i x i (2)
v(0) ! "'A sin ( ! v i (1)
x(0) ! A cos ( ! x i |