Answer:
x=±0.026m
Explanation:
In simple harmonic motion the maximum value of the magnitude of velocity
[tex]v_{max}=wA=\sqrt{\frac{k}{m} }A[/tex]
The speed as a function of position for simple harmonic oscillator is given by
[tex]v=w\sqrt{A^{2}-x^{2}[/tex]
where A is amplitude of motion
Given data
Amplitude A=3 cm =0.03 m
v=(1/2)Vmax
To find
We have asked to find position x does its speed equal half of is maximum speed
Solution
The speed of the particle the maximum speed as:
[tex]v=\frac{V_{max} }{2}\\ w\sqrt{A^{2}-x^{2} }=\frac{wA}{2}\\ A^{2}-x^{2}=\frac{A^{2} }{4}\\ x^{2}=A^{2}- \frac{A^{2} }{4}\\ x^{2}=\frac{3A^{2}}{4} \\x=\sqrt{\frac{3A^{2}}{4}}[/tex]
x=±(√3(0.03)/2)
x=±0.026m
The position at which the speed equals half of is maximum speed is ±0.026m or ±2.6cm
The instantaneous speed in the simple harmonic motion is given by:
[tex]v=\omega \sqrt{A^2-x^2}[/tex]
where ω is the angular frequency, A is the amplitude = 3cm, and x is the instantaneous displacement.
The maximum speed is given by [tex]v_{max}=\omega A[/tex]
Now, the position at which the speed becomes half the maximum speed can be calculated as follows:
[tex]\frac{v_{max}}{2} =\omega \sqrt{A^2-x^2}\\\\\frac{\omega A}{2}=\omega \sqrt{A^2-x^2} \\\\\frac{A^2}{4}=A^2-x^2\\\\x=\sqrt{\frac{3A^2}{4} }\\\\x=\sqrt{\frac{3\times(3\times10^{-2})^2}{4} }\\\\x=0.026m\\or\\x=-0.026m[/tex]
Learn more about simple harmonic motion:
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