Halliday

Chapter 15

9. um auto-falante produz um som musical através das oscilações de um diafragma, cuja amplitude é limitada por.
The magnitude of the maximum acceleration is given by am = 2xm, where  is the angular frequency and xm 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 greater than 498 Hz, the acceleration exceeds g for some part of the motion.

10. qual é a constante de fase do oscilador cuja função. (do gráfico)
We note (from the graph) that xm = 6.00 cm. Also the value at t = 0 is xo =  2.00 cm. Then Eq. 15-3 leads to

 = cos1(2.00/6.00) = +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 = 0.

11.a função x=(6m)cos[(3pi 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, we find

(c) Differentiating again, we obtain

(d) In the second paragraph after Eq. 15-3, the textbook defines the phase of the motion. In this case (with t = 2.0 s) the phase is 3(2.0) + /3  20 rad.

(e) Comparing with Eq. 15-3, we see that  = 3 rad/s. Therefore, f = /2 = 1.5 Hz.

(f) The period is the reciprocal of the frequency: T = 1/f  0.67 s.

12. qual é a constante de fase do oscilador harmônico cuja função (gráfico)
We note (from the graph) that vm = xm = 5.00 cm/s. Also the value at t = 0 is vo = 4.00 cm/s. Then Eq. 15-6 leads to

 = sin1(4.00/5.00) = – 0.927 rad or +5.36 rad.

The other “root” (+4.07 rad) can be rejected on the grounds that it would lead to a positive slope at t = 0.

13. na fig 15-31 duas molas iguais, de constante elástica 7508 N/m. (mola~bloco~mola).
When displaced from equilibrium, the net force exerted by the springs is –2kx