First use the results discussed in the textbook to calculate the angular frequen

Question

First use the results discussed in the textbook to calculate the angular frequency: omega = (k/m)^1/2 = (7 N/m/2.18 times 10^-1 kg)^1/2 = 5.7 rad/s. Then use the relation between period and angular frequency to find the period: T = 2 pi/omega = 1.1 s. (B) Determine the maximum speed of the block. Recall from the textbook that x = A cos(omega t). where A, the maximum displacement, is equal to 6 cm, the initial displacement. The velocity is given by v = - A omega sin (omega t). So the maximum speed occurs when the sine function has the value plusminus 1, where its speed is |v_max| = omega A = (5.7 rad/s)(0.06 m) = 0.3422. m/s. (C) What is the maximum acceleration of the block? The acceleration is given by a = -A omega^2 cos(omega t). The acceleration has its largest magnitude when the cosine function is plusminus 1, where | a_max| = omega^2 A = (5.7 rad/s)^2 (0.06 m) = 1.95 m/s^2. (D) Express the position, speed, and acceleration as functions of time. Using the expressions for position, speed, and acceleration, we find the following: x = A cos(5.7t) = (0.06 m cos(5.7t) v = - omega A sin(5.7t) = 0.342 m/s) sin(5.7 t), a = - omega^2 cos(5.7t) = 1.95 m/s^2 cos (5.7t). FINALIZE Notice that when the displacement x is at a maximum, at x = A or x = -A, the velocity is zero, and when x is zero, the magnitude of the velocity is a maximum. Further, when x = +A, its most positive value, the acceleration is a maximum but in

Explanation / Answer

the numerical values are correct. just put the negative sign in front of these values.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.