Question

A mass of 0.5 kg is attached to the end of a massless spring of spring constant 0.40 N/m. It is released from rest from an extended position. After 0.6 s, the speed of the mass is measured to be 2.5 m/s. What is the amplitude of oscillation?

What is the total energy (relative to the mass at rest in the unextended position) contained in this system?

Answer 1

The angular frequency ω = (k/m)^1/2

Where

spring constant k = 0.40 N/m

Mass m = 0.5 kg

Hence ω = (0.40/0.5)^1/2

ω = 0.89 /s

Also From the equation of velocity for simple harmonic oscillation

v = - A ω Cos ωt

At 0.6 s the speed is 2.5 m/s

Hence

- A = v/ ω cos ωt
​- A = 2.5 / (0.89) cos (0.89)(0.6)

- A = 3.26 m

This gives the amplitude in negative direction.

In both the directions the maximum displacement or Amplitude , A = 3.26 m

At Unextended position or equilibrium position, The kinetic energy is maximum and potential energy is zero.

Hence the total energy is

E = ½ m (v_max)²

E = ½ m (A ω)²

   = ½ (0.5) ( 3.26 * 0.89)²

E = 2.1 Joules