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  • Amplitude of harmonic oscillator is b when velocity of the particle is 1 by 4 of maximum velocity then determine position of particle  Option: 1</str

Amplitude of harmonic oscillator is b when velocity of the particle is 1 by 4 of maximum velocity then determine position of particle
 

Option: 1

\frac{4}{\sqrt {15}}. B


Option: 2

\frac{\sqrt {15}}{2}. B


Option: 3

\frac{\sqrt {15}}{4}. B


Option: 4

\frac{3}{4}.B


Answers (1)

best_answer

Amplitude = B

Vmax = Bw

VP = \frac{V_{max}}{4}  x= ?

If we know V as a fn of x.

\begin{aligned} & \frac{V^2}{(B w)^2}+\frac{x^2}{B^2}=1 \\ & \frac{v^2}{B^2 w^2}+\frac{x^2}{B^2}=1 \\ & \frac{1}{B^2}\left[\frac{v^2}{w^2}+\frac{x^2}{1}\right]=1 \\ & \frac{v^2}{w^2}+\frac{x^2}{1}=B^2 \\ & \frac{v^2+x^2 w^2}{w^2}=B^2 \\ & v^2+x^2 w^2=B^2 w^2 \\ & v^2=B^2 w^2-x^2 w^2 \\ & v^2=w^2\left(B^2-x^2\right) \end{aligned}\\\\ \begin{aligned} & \frac{V_{\max }}{4}=w \sqrt{B^2-x^2} \\\\ & \frac{B w}{4}=w{\sqrt{B^2-x^2}} \\\\ & \frac{B}{4}=\sqrt{B^2-x^2} \end{aligned}\\\\ \begin{aligned} & \frac{B^2}{16}=\frac{B^2-x^2}{1} \\ & B^2=16 B^2-16 x^2 \\ & 16 x^2=16 B^2-B^2 \\ & 16 x^2=15 B^2 \\ & x^2=\frac{15 B^2}{16} \\ & x=\sqrt{15 / 16 B^2} \\ & x=\frac{\sqrt{15}}{4} \cdot B \end{aligned}

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