Question

Which of the following correctly represents the equation of a simple harmonic oscillator?
d2xdt2+ωx=0
d2xdt2+ω2x2=0
dxdt2+ω2x=0
d2xdt2+ω2x=0

Answer 1

Answer on Question #41349 – Physics – Mechanics

Which of the following correctly represents the equation of a simple harmonic oscillator?

$$\begin{array}{l} \mathrm{d}2x\mathrm{d}t2 + \omega x = 0 \\ \mathrm{d}2x\mathrm{d}t2 + \omega 2x2 = 0 \\ \mathrm{d}x\mathrm{d}t2 + \omega 2x = 0 \\ \mathrm{d}2x\mathrm{d}t2 + \omega 2x = 0 \\ \end{array}$$

Solution:

For one-dimensional simple harmonic motion, the equation of motion, which is a second-order linear ordinary differential equation with constant coefficients, could be obtained by means of Newton's second law and Hooke's law.

$$F_{\text{net}} = m \frac{d^2 x}{dt^2} = -k x$$

where $m$ is the inertial mass of the oscillating body, $x$ is its displacement from the equilibrium (or mean) position, and $k$ is the spring constant.

Therefore,

$$\frac{d^2 x}{dt^2} = - \left(\frac{k}{m}\right) x$$

Substitution $\omega^2 = \frac{k}{m}$:

$$\frac{d^2 x}{dt^2} + \omega x = 0$$

Hence, the correct answer is first: $\mathrm{d}2x\mathrm{d}t2 + \omega x = 0$

Answer: $\mathrm{d}2x\mathrm{d}t2 + \omega x = 0$

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