JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2021
- Medium
- Numerical
A particle executes simple harmonic motion represented by displacement function as \(x(t)=A \sin (\omega t+\phi)\) If the position and velocity of the particle at \(t=0\, {s}\) are \(2\, {cm}\) and \(2\, \omega \,{cm} \,{s}^{-1}\) respectively, then its amplitude is \(x \sqrt{2} \,{cm}\) where the value of \(x\) is ..... .
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Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
As given
\(v(t)=A \omega \cos (\omega t+\phi)\)
\(2=A \sin \phi.....(1)\)
\(2 \omega=A \omega \cos \phi.....(2)\)
From \((1)\) and \((2)\)
\(\tan \phi=1\)
\(\phi=45^{\circ}\)
Putting value of \(\phi\) in equation \((1)\)
\(2=A\left\{\frac{1}{\sqrt{2}}\right\}\)
\(A=2 \sqrt{2}\)
\(x=2\)
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