An undamped 2.65 kg horizontal spring oscillator has a spring constant of 38.5 N/m. While oscillating, it is found to have a speed of 2.92 m/s as it passes through its equilibrium position. What is its amplitude of oscillation?

Answer: 0.44831m

Explanation:

Unwanted horizontal spring oscillator=2.65kg

Spring constant = 38.5N/M

Speed=2.92m/s

Amplitude of oscillation=?

Potential energy=m*v2/2

=2.65*2.92/2

=3.8695J

Potential energy=kinetic energy

Potential energy=1/2kx^2

3.869=1/2*38.5*x^2

3.869=19.25x^2

Dividing both sides by 19.25

3.869/19.25=x^2

So therefore, x^2=√0.200987

x=0.44831m