JEE MAIN - Physics (2009 - No. 18)
If $$x,$$ $$v$$ and $$a$$ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $$T,$$ then, which of the following does not change with time?
$$aT/x$$
$$aT + 2\pi v$$
$$aT/v$$
$${a^2}{T^2} + 4{\pi ^2}{v^2}$$
Giải thích
For an $$SHM,$$ the acceleration $$a = - {\omega ^2}x$$ where $${\omega ^2}$$ is a constant. Therefore $${a \over x}$$ is a constant. The time period $$T$$ is also constant. Therefore $${{aT} \over x}$$ is a constant.