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  • A Particle executes S.H.M with amplitude ' A ' and time period ' T ' the displacement of the Particle When its speed is half of maximum speed is  <img alt="\frac

A Particle executes S.H.M with amplitude ' A ' and time period ' T ' the displacement of the Particle When its speed is half of maximum speed is  \frac{\sqrt{x}}{2} \cdot A  find the value of  A  


 

Option: 1

A=\frac{6 x}{\sqrt{3}}


Option: 2

A=\frac{2 x}{\sqrt{3}}


Option: 3

A=\frac{4 x}{\sqrt{3}}


Option: 4

A=\frac{3 x}{\sqrt{3}}


Answers (1)

best_answer

V=\frac{V_{\max }}{2} \\
w \sqrt{A^2-x^2}=\frac{A \omega}{2} \\
A^2-x^2=\frac{A^2}{4} \\
\frac{x}{1}=\frac{\sqrt{3}}{2} \cdot A \\

2 x=\sqrt{3} \cdot A \\

A=\frac{2 x}{\sqrt{3}}

Posted by

manish painkra

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