Question

A 472-g glider on a horizontal, frictionless air track is attached to an ideal spring whose force constant is 712 N/m. At one point in time, the glider is moving at 0.49 m/s and is 9.16 cm from the equilibrium point. What is the amplitude of motion

Answer 1

0.09 m

Step-by-step explanation:

The best way to solve this would be by comparing energy.

Build conservation of energy formula.

`K+U=E\\`

Find kinetic energy:

`1/2mv^2=K\\1/2*0.472*0.49^2=K\\0.057=K`

Now find the potential spring energy at 9.16cm:

`1/2KA^2=Us\\1/2*712*0.0916^2=Us\\2.987 = Us`

Since all energy must come from potential energy, we can substitute values:

`0.057+2.987=1/2kA^2\\3.04400=712/2*A^2\\0.008=A^2\\0.089=A\\0.09=A`