Question

An object spring system moving with simple harmonic motion has amplitude A. When the kinetic energy of the object equals twice the potential energy stored in the spring, what is the position x of the object? (a)$A$ (b)$\frac{1}{3}A$ (c)$\frac{A}{\sqrt{3}}$ (d) 0 (e) none of those answers .

Short answer

Option (c) is correct. The position x of the object is $x=\frac{A}{\sqrt{3}}$.

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 position x of the object

The total energy of the object spring system is

$\frac{1}{2}k{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$

When the kinetic energy is twice the potential energy,

$\frac{1}{2}m{v}^2=2\left(\frac{1}{2}k{x}^2\right)=k{x}^2$, and the total energy is

$$\begin{gathered} \frac{1}{2}k{A}^2=k{x}^2+\frac{1}{2}k{x}^2 \\ \frac{1}{2}k{A}^2=\frac{3}{2}k{x}^2 \\ x=\frac{A}{\sqrt{3}} \end{gathered}$$

Option (c) is correct. The position x of the object is $x=\frac{A}{\sqrt{3}}$.