Answer to Question #198040 in Mechanics | Relativity for Jamarlon Williams

Question

A particle executes simple harmonic motion with an amplitude of 4.72 cm. At what positive displacement from the midpoint of its motion does its speed equal one half of its maximum speed?

Answer 1

For an object in simple harmonic motion, the maximum value of the magnitude of velocity:

$v_{max}=ωA = \sqrt{ \frac{k}{m} }A$

The speed as a function of position for a simple harmonic oscillator is given by:

$v = ω \sqrt{A^2-x^2}$

where A is the amplitude of the motion.

$A = 4.72 \;cm = 0.0472 \;m$

The speed of the particle is half the maximum speed:

$v = \frac{1}{2}v_{max} \\ ω \sqrt{A^2-x^2} = \frac{ωA}{2} \\ \sqrt{A^2-x^2}=\frac{A}{2} \\ A^2-x^2 = \frac{A^2}{4} \\ x^2 = A^2 -\frac{A^2}{4} \\ = \frac{3A^2}{4} \\ x = \sqrt{ \frac{3A^2}{4} } \\ = ±\frac{\sqrt{3}A}{2} \\ x = ±\frac{\sqrt{3} \times 0.0472}{2} \\ = ±0.0408\;m$