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RSB [31]
3 years ago
13

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 spe

ed 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?
Physics
1 answer:
stich3 [128]3 years ago
4 0

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

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