# ch15 - 641 Chapter 15 1(a During simple harmonic motion the...

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Unformatted text preview: 641 Chapter 15 1. (a) During simple harmonic motion, the speed is (momentarily) zero when the object is at a “turning point” (that is, when x = + x m or x = – x m ). Consider that it starts at x = + x m and we are told that t = 0.25 second elapses until the object reaches x = – x m . To execute a full cycle of the motion (which takes a period T to complete), the object which started at x = + x m , must return to x = + x m (which, by symmetry, will occur 0.25 second after it was at x = – x m ). Thus, T = 2 t = 0.50 s. (b) Frequency is simply the reciprocal of the period: f = 1/ T = 2.0 Hz. (c) The 32 cm distance between x = + x m and x = – x m is 2 x m . Thus, x m = 32/2 = 16 cm. 2. (a) The acceleration amplitude is related to the maximum force by Newton’s second law: F max = ma m . The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is a m = 2 x m , where is the angular frequency ( = 2 f since there are 2 radians in one cycle). The frequency is the reciprocal of the period: f = 1/ T = 1/0.20 = 5.0 Hz, so the angular frequency is = 10 (understood to be valid to two significant figures). Therefore, = = 0.12 10 0.085 = 10 . 2 2 F m x m max kg rad / s m N b gb gb g (b) Using Eq. 15-12, we obtain 2 2 2 0.12kg 10 rad/s 1.2 10 N/m. k k m m 3. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is a m = 2 x m , where is the angular frequency ( = 2 f since there are 2 radians in one cycle). Therefore, in this circumstance, we obtain 2 2 2 2 (2 ) 2 6.60 Hz 0.0250 m 43.0 m/s . m m m a x f x 4. (a) Since the problem gives the frequency f = 3.00 Hz, we have = 2 f = 6 rad/s (understood to be valid to three significant figures). Each spring is considered to support one fourth of the mass m car so that Eq. 15-12 leads to 2 5 car 1 1450kg 6 rad/s 1.29 10 N/m. / 4 4 k k m CHAPTER 15 642 (b) If the new mass being supported by the four springs is m total = [1450 + 5(73)] kg = 1815 kg, then Eq. 15-12 leads to 5 new new total 1 1.29 10 N/m 2.68Hz. / 4 2 (1815/ 4) kg k f m 5. (a) The amplitude is half the range of the displacement, or x m = 1.0 mm. (b) The maximum speed v m is related to the amplitude x m by v m = x m , where is the angular frequency. Since = 2 f , where f is the frequency, 3 = 2 = 2 100 Hz 1.0 10 m = 0.63 m/s. m m v fx (c) The maximum acceleration is 2 2 2 3 2 2 = = 2 = 2 100 Hz 1.0 10 m = 3.9 10 m/s ....
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ch15 - 641 Chapter 15 1(a During simple harmonic motion the...

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