Question

A block spring system vibrating on a frictionless, horizontal surface with amplitude of 6.0 cm has energy of 12 J. If the block is replaced by one who’s mass is twice the mass of the original block and the amplitude of the motion is again 6.0 cm, what is the energy of the system? (a) 12 J (b) 24 J (c) 6 J (d) 48 J (e) none of those answers

Short answer

The spring constant k and the amplitude A remain unchanged, the total energy is unchanged. So, option (a) will be the correct answer for this question.

Step-by-step solution

Step 1: The total energy of a simple harmonic oscillator

The total energy of a simple harmonic oscillator is a constant of the motion and is given by

$E=\frac{1}{2}k{A}^2$

$E=$Total energy

$L=$Length of pendulum

$K=$Constant

Find the energy of the system

In this spring mass system, the total energy equals the elastic potential energy at the moment the mass is temporarily at rest at $x=A=6\text{cm}$(i.e., at the extreme ends of the simple harmonic motion). Thus, $E=\frac{1}{2}k{A}^2$ and we see that as long as the spring constant k and the amplitude A remain unchanged, the total energy is unchanged.