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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
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The phase constant in radians of a simple pendulum having an initial displacement of 4 units and having an amplitude of 3 is

$\begin{array}{1 1}(a)\;0.375 \\(b)\;1.33 \\( c)\; 1 \\(d)\;0 \end{array}$
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$x= 3 \sin wt +\phi$
At $t=0;$
$x= 4= 3 \sin \phi$
$\therefore \phi$ in radius $=\large\frac{4}{3}$
$(\sin \theta=\theta$ in radians)
Hence b is the correct answer
answered Dec 11, 2013 by meena.p

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