Question

A 0.28 kg mass is attached to a light spring with a force constant of 33.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.0 cm and released from rest, determine the following. (a) maximum speed of the oscillating mass m/s (b) speed of the oscillating mass when the spring is compressed 1.5 cm m/s (c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position m/s (d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value m

Answer 1

Answer:

Explanation:

ω = √k/m = √(33.9/0.28) = 11 rad/s

(a) maximum speed of the oscillating mass

vmax = ωA = 11(0.05) = 0.55 m/s

(b) speed of the oscillating mass when the spring is compressed 1.5 cm

The portion of total energy that is not spring potential is kinetic

½kA² - ½kx² = ½mv²

v = √(k(A² - x²)/m) = √(33.9(0.05² - 0.015²)/0.28 = 0.52482... ≈ 0.52 m/s

(c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position

Different wording, but same question as part (b)  0.52 m/s

(d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value m

The portion of total energy that is not kinetic is spring potential

½kA² - ½mv² = ½kx²

x = √(kA² - m(vmax/2)²) / k) = √(33.9(0.05²) - 0.28(0.55/2)²) / 33.9)

x = 0.043305...≈ 4.3 cm