JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 22)
Time period of a simple pendulum is T. The time taken to complete $${5 \over 8}$$ oscillations starting from mean position is $${\alpha \over \beta }T$$. The value of $$\alpha$$ is _________.
Answer
7
Explanation
_26th_February_Evening_Shift_en_22_2.png)
$${5 \over 8}$$ oscillation = $${1 \over 2}$$ oscillation + $${1 \over 8}$$ oscillation
From figure, A to B = $${1 \over 2}$$ oscillation and B to C is $${1 \over 8}$$ oscillation.
$$ \therefore $$ $$\pi { + \,\theta = \omega t} $$
$$ \Rightarrow $$ $$\pi { + {{\pi{} } \over 6}} = \omega t$$
$$ \Rightarrow $$ $${{7\pi {} } \over 6} = \left( {{{2\pi {} } \over T}} \right)t$$
$$ \Rightarrow $$ t = $${{7T} \over {12}}$$
Comments (0)
