Question

A 2.00 kg frictionless block is attached to a horizontal spring as shown. Spring constant k = 200.00 N/m. At t = 0, the position x = 0.225 m, and the velocity is 4.25 m/s toward the right in the positive x direction. Position x as a function of t is: x = A*cos(?t + theta) , where A is the amplitude of motion and ? is the angular frequency discussed Chapter 11 and the notes. Theta is called the phase constant that will be addressed for extra credit below. (a) Use conservation of energy to compute amplitude A. (b) How much farther from the point shown will the block move before it momentarily comes to rest before turning around? (c) What is the period T of the motion? (d) If the mass of this problem was doubled to 4.00 kg, how would your answer to part (c ) change? (e) Use your trigonometry background to find phase constant theta.

t=0 +4.25 mis =0 0.225 m

Answer 1

Solution Energy at t-0is 0.5(2 kg) (4.25 m/s)0.5(200 N/m) (0.225 m) -23.13 J When this energy is completely converts into potential energy will be Amplitude 0.5(200 Nm)A 23.13J 100 N/m xA-23.13 J 2 23.13 J 100 N/m 23.13 J 100 N/m -0.48 nm Hence the amplitude is 0.48 m

It will go up to 0.48 m is calculated as (0.48 m-0.225 m) - 0.26 m Hence the position of the block is 0.261m 200 N/m (2 kg) 10 rad/s 2π 6.28 10 0.628 s Hence the period of the motion is 0.628 s

If the mass is doubled 4 kgmeans angular velocity becomes - So that, 1.414x0.628 -0.88 s Hence the mass is doubled 4 kg means period of the motion is 0.88 s At initial time t -0 0.225-0.48 cos θ 0.225 0.48 cos θ 0.468 θ-cos-1 (0.468 62.04o Hence the phase constant is 62.04°