Question

In a lab, students connect a cart with a mass of 1.70 kg to a spring of equilibrium length 55.8 cm and spring constant 31.2 N/m. The students pull the cart to the right until the spring has a length of 87.3 cm, as shown in position 1, then release it from rest. The cart travels through the three positions which are just released, at equilibrium, and furthest left. Immediately after it is released the cart will have an acceleration of magnitude of how many m/s^2? Question two. As it travels to the left the carts maximum speed will be m/s. Question 3. The cart will take how many seconds to return to where it was released? Question 4. In a second expirèrent, the students pull the cart further back before relapsing it. Compared to the first expirèrent the time it takes the cart to return to where it was relaesed will be Less, greater or the same?

Answer 1

1) F = - K X where F is the restoring force and K the spring constant

F = 31.2 N/m * (.873 - .558) m = 9.83 N

a = F / m = 9.83 / 1.70 = 5.78 m/s^2

2) ω = (K / m)^1/2 = (31.2 / 1.70)^1/2 = 4.28 / s

Vmax = ω A = 4.28 / s * (.873 - .558) m = 1.35 m/s

(note also a = ω^2 A = 4.28^2 * (.873 - .558) = 5.77 the result of (1))

(3) f = ω / (2 * π) = 4.28 / (2 * 3.14) = .681 / s frequency of vibration

P = 1 / f = 1 / .681 = 1.47 s period of oscillation

P / 4 = .367 s time for 1/4 oscillation

(4) Since ω is constant for the conditions given the time to return to the origin does not depend on the amplitude of oscillation