14 The objects then given a slight push u tes SHM. the period Tof this oscillat

Question

14 The objects then given a slight push u tes SHM. the period Tof this oscillat stretches a vertical spring 0215 m. I stretched an additional 0.130 m and rel it take to reach the (new) equilibrium spring is on again? n650 kg mass oscillates according to the equation 40t) x is meters and t seconds. Determine (a) the amplitude (b) frequency, (c) the total energy, and (d the kinetic energy and potential energy when x 0.360 m 0.25 kg mass at the end of a spring oscillates 2.2 15 (l) A times per second with an amplitude of 0.15 m. Determine (a) the speed when it passes the equilibrium point, (b) speed when it is 0.10 from the total energy of the system, and (d) the equation describing the motion of the mass, assuming that at t 0, x was a l6 (T) It takes a force of 91.0 N to compress the spring of a toy popgun 0.175 m to load" a 0.160-kg ball. With what speed will the ball leave the gun if fired horizontally? 17. (I) If one oscillation has 30 times the energy of a second one of equal frequency and mass, what is the ratio of their amplitudes? l (l) A mass of 240 g oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4Sam. (a) at is the effective spring constant for this motion? (b) How much energy is involved in this motion? A mass resting on a horizontal, frictionless surface is ched to one end of a spring: the other end is fixed to all. It takes 3.6 J of work to compress the spring by from the spring is compressed, and the released rest, it expe acceleration mass hind the value of (a) the spring constant the (I) A ject with mass 2.7 is executing simple har- kg constant motion, attached to a spring with spring equi um the object is 0.020 m its m/s. ion, it is moving with a ed of Calculate amplitude of the spee (b) maximum speed attaine m. by the object. 0, at rest on the end of a an 885-g mass by a hammer spring (k

Explanation / Answer

16) Potential energy = 0.5 * 91 * 0.175 = 7.96

Kinetic energy = 1/2 * 0.160 * v2 = 7.96

v = 9.97 m/s

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17) E2 = 3 E1

E = 1/2 * k A2 * sin2 (wt)

E1 = 1/2 * k A12 * sin2 (wt)

E2 = 1/2 *k A22 * sin2 (wt)

E2 / E1 = 3 = (A2/A1)2

(A2/A1) = sqrt(3) = 1.73

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