JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 16)
A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $${1 \over a}$$s. The value of 'a' to the nearest integer is _________.
Respondre
6
Explicació
Time period (T) = 2 sec.
X = A sin ($$\omega$$t + $$\phi$$) ($$\phi$$ = 0 at M.P.)
$$ \Rightarrow $$ $${A \over 2} = A\sin {{2\pi } \over T}t$$
$$ \Rightarrow $$ $${{2\pi } \over 2}t = {\pi \over 6}$$
$$ \Rightarrow $$ $$t = {1 \over 6}$$
$$ \therefore $$ a = 6
X = A sin ($$\omega$$t + $$\phi$$) ($$\phi$$ = 0 at M.P.)
$$ \Rightarrow $$ $${A \over 2} = A\sin {{2\pi } \over T}t$$
$$ \Rightarrow $$ $${{2\pi } \over 2}t = {\pi \over 6}$$
$$ \Rightarrow $$ $$t = {1 \over 6}$$
$$ \therefore $$ a = 6