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The phase constant in radians of a simple pendulum having an initial displacement of $3.9$ units and having an amplitude of $5.2$ is

$\begin{array}{1 1}(a)\;0.375 \\(b)\;0.75 \\(c)\; 1 \\(d)\;0 \end{array}$

1 Answer

$x= 5.2 \sin wt +\phi$
At $t=0;$
$x= 3.9= 5.2 \sin \phi$
$\therefore \phi$ in radius $=\large\frac{3.9}{5.2}$
$(\sin \theta=\theta$ in radians)
Hence b is the correct answer
answered Dec 10, 2013 by meena.p

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