A spring is stretched 5 cm by a force of 50 dynes. A mass of 10 g is placed on t

Question

A spring is stretched 5 cm by a force of 50 dynes. A mass of 10 g is placed on the lower end of the spring. After equilibrium has been reached, the upper end of the spring is moved up and down so that the extrenal force acting on the mass is given by F(t) = 20cos(wt), t>0. If the damping force is equal to twice the speed (i.e. B = 2 m/s). (a) Find the position of the mass at any time, measured from its equilibrium position. (b) Find the value of w for which resonance occurs. (c) The amplitude at a driving frequency, w = 0.5 rad/s.

Explanation / Answer

F = force applied to spring = 50 dynes = 50 x 105 N

x = strecth in the spring caused = 5 cm = 0.05 m

k = spring constant

Using the equation

F = k x

50 x 105 = k (0.05)

k = 1 x 108 N/m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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