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;
[tex]v = \omega \sqrt{A^2-x^2}[/tex]
the maximum speed of the simple harmonic motion is given as;
[tex]v_{max} = \omega A[/tex]
when the speed equal one fourth of its maximum speed
[tex]v =\frac{v_{max}}{4}[/tex]
[tex]\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[/tex]
Thus, the positions are 0.0194 m and - 0.0194 m.
The positions where the speed equals 1/4 of its maximum speed is mathematically given as
[tex]x \pm 0.0194[/tex]
Question Parameters:
an amplitude of 2.00 cm.
Generally, the equation for the speed of simple harmonic motion is mathematically given as
[tex]v = w \sqrt{A^2-x^2}\\\\v=\frac{wA}{4}[/tex]
Therefore
[tex]w \sqrt{A^2-x^2}=\frac{wA}{4}\\\\x^2=A^2-A^2/16\\\\x=\sqrt{\frac{15A^2}{16}}[/tex]
[tex]x \pm 0.0194[/tex]
In conclusion, the positions are
[tex]x \pm 0.0194[/tex]
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