Question

A 0.253 kg mass is attached to a spring.
The mass is given a small tug, causing it to oscillate.
The observed motion of the mass is given by the following equation:
x(t) = 0.0289 m * cos(9.09 rad/s * t).

What is the amplitude of motion? 0.0289 m

What is the frequency of motion? 1.45 Hz

What is the spring constant? 20.9 N/m

What is the total energy of the system? 0.00873 J

What is the kinetic energy of the mass at time 1.51 s?

What is the potential energy of the spring at time 1.51 s?

Answer 1

Kinetic energy is given by 0.5*m*v²

m=0.253 kg

V=dx/dt=-0.0289*9.09*sin(9.09*t)=-0.2627*sin(9.09*t)

At t=1.51 s,

Velocity=-0.2827*sin(9.09*1.51)=-0.259 m/s

Kinetic energy=0.5*0.253*0.259²=0.0085 J

Potential energy=0.5*k*x²

K=20.9 N/m

x=0.0289*cos(9.09*1.5)=0.0116 m

Potential energy=0.5*k*x²=0.5*20.9*0.0116²=0.0014 J