Chapter15

# Chapter15 - 1 The textbook notes(in the discussion...

This preview shows pages 1–8. Sign up to view the full content.

1. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is a m = ω 2 x m , where is the angular frequency ( = 2 π f since there are 2 π radians in one cycle). Therefore, in this circumstance, we obtain () 2 22 2 (2 ) 2 6.60 Hz 0.0220 m 37.8 m/s . mm m ax f x ωπ π == = =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2. (a) The angular frequency ω is given by = 2 π f = 2 π / T , where f is the frequency and T is the period. The relationship f = 1/ T was used to obtain the last form. Thus = 2 π /(1.00 × 10 5 s) = 6.28 × 10 5 rad/s. (b) The maximum speed v m and maximum displacement x m are related by v m = x m , so == 1.00 10 6.28 10 =1.59 10 . 3 5 3 x v m m × × × m/s rad / s m
3. (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 = ω x m , where is the angular frequency. Since = 2 π f , where f is the frequency, () 3 = 2 = 2 120 Hz 1.0 10 m = 0.75 m/s. mm vf x ππ × (c) The maximum acceleration is ( ) 2 2 23 2 2 = = 2 = 2 120 Hz 1.0 10 m = 5.7 10 m/s . m ax f x ωπ π ××

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(b) Using Eq. 15-12, we obtain () ( ) 2 22 0.12kg 10 rad/s 1.2 10 N/m. k km m ωω π = ¡ == = × 4. (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 = ω 2 x m , where is the angular frequency ( = 2 π f since there are 2 π 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 π (understood to be valid to two significant figures). Therefore, = = 0.12 10 0.085 = 10 . 2 2 Fm x m max kg rad / s m N b gb g b g π
5. (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 36 cm distance between x = + x m and x = – x m is 2 x m . Thus, x m = 36/2 = 18 cm.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6. (a) Since the problem gives the frequency f = 3.00 Hz, we have ω = 2 π f = 6 π 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 ωπ = ¡ == × (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 11 . 2 9 1 0 N / m 2.68Hz. / 4 2 (1815/ 4) kg k f m π × = ¡
7. (a) The motion repeats every 0.500 s so the period must be T = 0.500 s. (b) The frequency is the reciprocal of the period: f = 1/ T = 1/(0.500 s) = 2.00 Hz. (c) The angular frequency ω is = 2 π f = 2 π (2.00 Hz) = 12.6 rad/s.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

### Page1 / 113

Chapter15 - 1 The textbook notes(in the discussion...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online