Maximum acceleration of mass at the end of a spring
I have the question "A mass at the end of a spring oscillates with a period of $2.8s$. The maximum displacement of the mass from its equilibrium position is $16cm$. For this oscillating mass, Calculate its maximum acceleration."
From the previous questions I have worked out the amplitude to be $0.16m$ and the angular frequency to be $2.26$ rads$^{-1}$.
I have used the equation:
$a = w^2x$
Therefore, $a = (2.26$ rads$^{-1})^2 * 0.16 m$
Therefore, $a = 0.82 m/s^2$.
Is this correct ?
$\endgroup$ 21 Answer
$\begingroup$$a = x'' = -\omega^2 x$
Which is a second order differential equation with solution.
$x = A \sin (\omega t + \phi)$
There are other ways to write it, but this one is common. The phase shift isn't particularly relevant here.
Period $=\frac {2\pi}\omega = 2.8$
$a_{max} = \omega^2 A\\ (\frac {2\pi}{2.8})^2(0.16) \frac m{s^2}$
$\endgroup$ 6
