Question

A particle executes simple harmonic motion with an amplitude of 2.00 cm. At what positions does its speed equal one fourth of its maximum speed?

Answer 1

Answer:

The positions are  0.0194 m  and - 0.0194 m.

Explanation:

Given;

amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m

speed of simple harmonic motion is given as;

$v = \omega \sqrt{A^2-x^2}$

the maximum speed of the simple harmonic motion is given as;

$v_{max} = \omega A$

when the speed equal one fourth of its maximum speed

$v =\frac{v_{max}}{4}$

$\omega\sqrt{A^2-x^2} = \frac{\omega A}{4} \\\\\sqrt{A^2-x^2}= \frac{A}{4}\\\\A^2-x^2 = \frac{A^2}{16} \\\\x^2 = A^2 - \frac{A^2}{16} \\\\x^2 = \frac{16A^2 - A^2}{16} \\\\x^2 = \frac{15A^2}{16} \\\\x= \sqrt{\frac{15A^2}{16} } \\\\x = \sqrt{\frac{15(0.02)^2}{16} }\\\\x = 0.0194 \ m \ \ or\ - 0.0194 \ m$

Thus, the positions are  0.0194 m and - 0.0194 m.