1 Simple harmonic motion displacement velocity = dx/dt acceleration = dv/dt LECTURE 3 CP Ch 14 CP449

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3 SHM and energy CP445

4 txVaKE PE 0A0 -  2 A 0½ k A 2 T / 4 T / 2 3T / 4 T What are all the values at times t = T/4, T/2, 3T/4, T ? 0 +A+A -A-A Problem 3.1

5 Problem 3.2 A spring is hanging from a support without any object attached to it and its length is 500 mm. An object of mass 250 g is attached to the end of the spring. The length of the spring is now 850 mm. (a) What is the spring constant? The spring is pulled down 120 mm and then released from rest. (b) What is the displacement amplitude? (c) What are the natural frequency of oscillation and period of motion? (d) Describe the motion on the object attached to the end of the spring. Another object of mass 250 g is attached to the end of the spring. (d) Assuming the spring is in its new equilibrium position, what is the length of the spring? (f) If the object is set vibrating, what is the ratio of the periods of oscillation for the two situations?

6 Problem 3.3 A 100 g block is placed on top of a 200 g block. The coefficient of static friction between the blocks is The lower block is now moved back and forth horizontally in SHM with an amplitude of 60 mm. (a) Keeping the amplitude constant, what is the highest frequency for which the upper block will not slip relative to the lower block? Suppose the lower block is moved vertically in SHM rather than horizontally. The frequency is held constant at 2.0 Hz while the amplitude is gradually increased. (b) Determine the amplitude at which the upper block will no longer maintain contact with the lower block.

7 Solution 3.3 Identify / Setup 2 FNFN FGFG m 1 a 1y = 0 F N = m g F f =  N =  m g m 1 = 0.1 kg m 2 = 0.2 kg  = 0.20 A 2 = 60 mm = 0.06 m max freq f = ? Hz SHM a max = A  2 = A(2  f) 2 = 4  2 f 2 A 1 2 FNFN FGFG m1m1 1

8 Execute (a) max frictional force between blocks F f =  m 1 g max acceleration of block 1 a 1max = F f / m 1 =  g max acceleration of block 2 a 2max = a 1max =  g SHM a 2max = 4  2 f 2 A f = 2 Hz  g = 4  2 f 2 A f =  [  g / (4  2 A)] =  [(0.20)(9.8)/{(4)(  2 )(0.06)}] Hz = 0.91 Hz

9 Execute (b) max acceleration of block 1 (free fall) a 1max = g max acceleration of block 2 a 2max = a 1max = g SHM a 2max = 4  2 f 2 A g = 4  2 f 2 A A = g / (4  2 f 2 ) = (9.8) / {(4)(  2 )(2 2 )} m = m Evaluate

10 Answers 3.2 (a) k = 7.0 N.m -1 (b) A = m (c) f 1 = 0.84 Hz T 1 = 1.2 s (d) SHM (f) T 2 / T 1 = 1.4