Halliday
Chapter 15 g. um auto-falante produz um som musical através das oscilações de um diafragma, cuja amplitude é Imitada por. The magnitude of the maximum acceleration is given by am = C]2xm, where 0 is the angular frequency and xrn is the amplitude. (a) The angular frequency for which the maximum acceleration is g is given by , and the corresponding frequency is given by (b) For frequencies gr g for some part of th 10. qual é a constant e gráfico) cceleration exceeds ar 2 Swipe to page ja função. (do
We note (from the graph) that xm = 6. 00 cm. Also the value at t- O is xo n 2. 00 cm. Then Eq. 15-3 leads toa +1. 91 rad or— 4. 37 rad. The other “root” (+4. 37 rad) can be rejected on the grounds that it would lead to a positive slope at t = O. 1 1 . a função rad/s)t + pi/3 rad) descreve o movimento harmônico simples de um corpo. (a) Making sure our calculator is in radians mode, we find (b) Differentiating with respect to time and evaluating at t 2 0 s paragraph after Eq. 15-3, the textbook defines the phase of the otion.
In this case (with t = 2. 0 s) the phase is 30(2. 0) + 0/3 C] 20 rad. (e) Comparing with Eq. 15-3, we see that C] = 30 rad/s. Therefore, f = – 1. 5 Hz. (f) The period is the reciproca’ of the frequency: T l/f C] 0. 67 s. 12. qual é a constante de fase do oscilador harmônico cuja função (gráfico) We note (from the graph) that vm C]xm 5. 00 cm/s. Also the value at t = O is vo = 4. 00 cm/s. Then Eq. 15-6 leads to a = sinC11(ClC14. OO/5. OO) – 0. 927 rad or 5. 36 rad. The other “root” (+4. rad) can be rejected on the grounds that it 13. na fig 15-31 duas molas iguais, de constante elástica 7508 N/ When displaced from equllibrium, the net force exerted by the springs is —2kx acting in a direction so as to return the block to its equilibrium position (x = O). Since the acceleration , Newton’s second law yields Substituting x = xm cos(üt + C]) and simplib,’ing, we find where n is in radians per unit time. Since there are 20 radians in a cycle, and frequency f measures cycles per second, we obtain