Question

A 200 g mass is attached to a horizontal spring and executes SHM making one oscillation every 2.0 seconds. At a certain time the block is at x=0.20 m and is moving with a velocity of -0.36 m/s. Determine: a. The spring constant of the spring b. The amplitude of oscillation c. The velocity of the mass when the mass is 1.5 cm from the equilibrium point

Hw #2 due Tuesday Jan 15 A 200 g mass is attached to a horizontal spring and executes SHM making one oscillation every 2.0 seconds. At a certain time the block is at x-0.20 m and is moving with a velocity of -0.36 m/s. Determine: a. The spring constant of the spring The amplitude of oscillation The velocity of the mass when the mass is 1.5 cm from the equilibrium point b. c.

Answer 1

(a) Angular velocity (w) = 2 pi/T = 2pi/2 = pi rad/s therefore w = squareroot k/m rightdoublearrow k = mw^12 spring constant (k) = 200 times 10^-3 times (3.14)^2 k = 1.972 N/m (b) v = w squareroot A^2 - x^2 -0.36 = 3.17 (squareroot A^2 -(0.2)^2) 0.013 = A^2 -(0.2)^2 A = squareroot 0.13 + 0.2^2 amplitude A = 0.23 m (c) velocity (v) = 3.4 squareroot 0.23^2 - 0.015^2 v = 0.72 m/s