hw01_sol

hw01_sol - 4(a The acceleration amplitude is related to the...

This preview shows pages 1–5. Sign up to view the full content.

(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

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

View Full Document
20. We note that the ratio of Eq. 15-6 and Eq. 15-3 is v / x = – ω tan( t + φ ) where = 1.20 rad/s in this problem. Evaluating this at t = 0 and using the values from the graphs shown in the problem, we find φ = tan 1 (– v o / x o ) = tan 1 ( + 4.00/(2 × 1.20) ) =1.03 rad (or –5.25 rad). One can check that the other “root” (4.17 rad) is unacceptable since it would give the wrong signs for the individual values of v o and x o .
22. Eq. 15-12 gives the angular velocity: 100 N/m 7.07rad/s. 2.00 kg k m ω == = Energy methods (discussed in §15-4) provide one method of solution. Here, we use trigonometric techniques based on Eq. 15-3 and Eq. 15-6. (a) Dividing Eq. 15-6 by Eq. 15-3, we obtain =+ v x t ωω φ tan b g so that the phase ( t + ) is found from () ( ) 11 3.415 m/s tan tan . 7.07 rad/s 0.129 m v t x ωφ −− §· += = ¨¸ ©¹ With the calculator in radians mode, this gives the phase equal to –1.31 rad. Plugging this back into Eq. 15-3 leads to0.129m cos( 1.31) 0.500m. mm xx =− ¡ = (b) Since t + = –1.31 rad at t = 1.00 s, we can use the above value of to solve for the phase constant . We obtain = –8.38 rad (though this, as well as the previous result, can have 2 π or 4 π (and so on) added to it without changing the physics of the situation). With this value of , we find x o = x m cos = – 0.251 m. (c) And we obtain v o = – x m ω sin = 3.06 m/s.

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

View Full Document
= 1 2 1 2 1000 500 225 f k m ππ == N/m kg Hz .
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/05/2011 for the course PHYSICS 133 taught by Professor Steigman during the Fall '09 term at Ohio State.

Page1 / 10

hw01_sol - 4(a The acceleration amplitude is related to the...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online