Question

Chapter 15, Problem 014 Your answer is partially correct. Try again A simple harmonic oscillator consists of a block of mass 3.20 kg attached to a spring of spring constant 140 N/m, when t-1.00 s, the position and velocity of the block are χ 0.195 m and v = 3.940 m/s. (a) What is the amplitude of the oscillations? what were the (b) position and (c) velocity of the block at t (a) Number To.626777 (b) Numbe (c) Number 0 s? Un Units

Answer 1

Given that Mass = 3.20 k g spring constant = 140 N/ m t = 1.00 s position = 0.195 m (x) velocity = 3.940 m/ s (v) Find a) what is the amplitude of the oscillations b) position o c) velocity of the block at t = 0 Equation of motion for simple harmonic motion is x = A x sin (w x t + u) velocity v = w x A x cos (w x t + u) = w x squareroot A^2 - x^2 A = Amplitude w = angular frequency = squareroot k/m = squareroot 140/ 3.20 = squareroot 4.375 = 2.091 rad/ s u = initial phase given at t = 1.005 x_1.0 = 0.195 m v_1.0 = 3.940 m/s v_1.0 = w x squareroot A^2 x 3.0^y 3.940^2 = 2.091^2 x (A^2 - 0.195^2) A^2 = 15.5236/4.372281 + 0.038025 3.55045 + 0.03802 squareroot 3.588 A = 1.89 m x_1.0 = A x sin (w x 6 + u) 0.195 = 1.89 x sin (2.091 x 1.0 + u) sin (2.091 + u) = 0.195/ 1.89 Initial phase u . sin^-1 (0.195/ 1.89) - 2.091 sin^- (0.103)