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Oscillations

# The phase constant in radians of a simple pendulum having an initial displacement of 4 units and having an amplitude of 3 is

$\begin{array}{1 1}(a)\;0.375 \\(b)\;1.33 \\( c)\; 1 \\(d)\;0 \end{array}$

$x= 3 \sin wt +\phi$
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