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13.Oscillations
medium
A simple pendulum is set up in a trolley which moves to the right with an acceleration a on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle $\theta $ with the vertical
A
${\tan ^{ - 1}}\frac{a}{g}$ in the forward direction
B
${\tan ^{ - 1}}\frac{a}{g}$ in the backward direction
C
${\tan ^{ - 1}}\frac{g}{a}$ in the backward direction
D
${\tan ^{ - 1}}\frac{g}{a}$ in the forward direction
Solution
(b)In accelerated frame of reference, a fictitious force (pseudo force) ma acts on the bob of pendulum as shown in figure.
Hence, $\tan \theta = \frac{{ma}}{{mg}} = \frac{a}{g}$
$ \Rightarrow \theta = {\tan ^{ – 1}}\left( {\frac{a}{g}} \right)$ in the backward direction.
Standard 11
Physics
