JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 9)
A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -
$${{4\pi } \over 3}$$
$${3 \over 8}\pi $$
$${7 \over 3}\pi $$
$${{8\pi } \over 3}$$
Explanation
$$v = \omega \sqrt {{A^2} - {x^2}} \,\,$$ . . .(1)
$$a = - {\omega ^2}x$$ . . .(2)
$$\left| v \right| = \left| a \right|$$ . . .(3)
$$\omega \sqrt {{A^2} - {x^2}} = {\omega ^2}x$$
$${A^2} - {x^2} = {\omega ^2}{x^2}$$
$${5^2} - {4^2} = {\omega ^2}\left( {{4^2}} \right)$$
$$ \Rightarrow \,\,\,3 = \omega \times 4$$
$$T = 2\pi /\omega $$
$$a = - {\omega ^2}x$$ . . .(2)
$$\left| v \right| = \left| a \right|$$ . . .(3)
$$\omega \sqrt {{A^2} - {x^2}} = {\omega ^2}x$$
$${A^2} - {x^2} = {\omega ^2}{x^2}$$
$${5^2} - {4^2} = {\omega ^2}\left( {{4^2}} \right)$$
$$ \Rightarrow \,\,\,3 = \omega \times 4$$
$$T = 2\pi /\omega $$
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