HOW DO YOU DO THIS PHYSICS PROBLEM

A bunch of grapes is placed in a spring scale at a supermarket. The grapes oscillate up and down with a period of 0.43 s.


What is the maximum speed of the grapes if their amplitude of oscillation is 2.7 cm?
Vmax=_____m/s




thanks

The system approximates a simple harmonic oscillator, so we'll start with the period equation:

T = 2π√(m/k)

Where m is the mass of the grapes (assuming the spring is massless) and k is the spring constant. Since we don't know either of these values, we cannot completely solve this problem yet, but we can get a ratio:

0.43 s = 2π√(m/k)

0.0684 s = √(m/k)

m/k = 0.00468 s²

So the mass to spring constant ratio is 0.00468 s².

m = (0.00468 s²)k

We need to consider the potential energy stored in the spring when it is stretched or compressed to its maximum (2.7 cm or 0.027 cm).

Us = 1/2 kx²

The grapes will have their maximum velocity when the spring passes its equilibrium point (x = 0), at which point:

K = 1/2 mv²

So...

1/2 kx² = 1/2 mv²

Keeping in mind that m = (0.00468 s²)k

1/2 k(0.027 m)² = 1/2 (0.00468 s²)k v²

(3.645e-4 m²)k = (0.00234 s²)k v²

Divide k out of both sides:

(3.645e-4 m²) = (0.00234 s²) v²

v² = 0.156 m²/s²

v = 0.395 m/s

I hope that helps. Good luck!