Question

A 2 kg object undergoes simple harmonic motion with an amplitude of 0.16 m and a maximum acceleration of 5 m/s2 . Part A What is its energy?

Answer 1

Given that :

mass of the object, m = 2 kg

amplitude, A = 0.16 m

maximum acceleration, a_max = 5 m/s²

Part-A : its energy will be given as -

E = (1/2) m v_max²                                               { eq.1 }

where, v_max = maximum velocity = A

then E = (1/2) m A²²    { eq.2 }

using an equation, to find the value of '' :

a_max = A ²

Or ² = a_max / A                                                                  { eq.3 }

inserting the value of ² in eq.2, we get

E = (1/2) m A² (a_max / A)

E = (1/2) m A a_max                                                                      { eq.4 }

inserting the values in eq.4,

E = (0.5) (2 kg) (0.16 m) (5 m/s²)

E = 0.8 J