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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
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The phase constant in rad of a simple pendulum whose amplitude of oscillation is 2; initial displacement is 1, initial velocity is 4 units/sec, frequency in rad/sec is 6 is:

$\begin{array}{1 1}(a)\;1 \\(b)\;\frac{1}{2}-\frac{1}{3} \\( c)\; \frac{1}{2}+\frac{1}{3} \\(d)\;0 \end{array}$
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In the $\large\frac{1}{2}$ is due to initial displacement and its velocity amplitude being $Aw= 2 \times 6 =12 units /sec$ and thus a phase constant of $\large\frac{4}{12}$ ie , $\large\frac{1}{3}$ comes due to initial velocity and is negative.
Hence b is the correct answer.
answered Dec 11, 2013 by meena.p

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