Question

An undamped 1.12-kg horizontal spring oscillator has a spring constant of 32.3 N/m. While oscillating, it is found to have a speed of 3.58 m/s as it passes through its equilibrium position. What is its amplitude of oscillation? What is the oscillator\'s total mechanical energy as it passes through a position that is 0.633 of the amplitude away from the equilibrium position?

Answer 1

Answer

given,

mass of spring, m = 1.12 Kg

spring constant , k = 32.3 N/m

speed. v = 3.58 m/s

a) Kinetic energy of the spring

$KE = \dfrac{1}{2}kA^2$

the mechanical work done by the spring

$KE = \dfrac{1}{2}mv^2$

now,

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2$

$A^2 = \dfrac{mv^2}{k}$

$A^2= \dfrac{1.12\times 3.58^2}{32.3}$

     A =0.666 m

Amplitude of Oscillation =  0.666 m

b) Total mechanical work done by the spring

  $E = \dfrac{1}{2} m v^2$

  $E = \dfrac{1}{2}\times 1.12 \times 3.58^2$

          E = 7.17 J