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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 ?

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    An `awesome score = 606` should mean you can type using mathjax. Please use that, it will garner more interest in your questions if you have nice formatting. – 2017-01-05
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    Okay thanks for the advice (: – 2017-01-05

1 Answers 1

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$a = x'' = -\omega^2 x$

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 $=\frac {2\pi}\omega = 2.8$

$a_{max} = \omega^2 A\\ (\frac {2\pi}{2.8})^2(0.16) \frac m{s^2}$

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    The period is actually $\frac{2\pi}{\omega}$ – 2017-01-05
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    Right right, brain farts. – 2017-01-05
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    The angular frequency (w) I have worked out on the previous question to be 2.26 rads$^-1$. So should it not be (2.26)$^2$ x 0.16 ? If we are using a = w$^2$ A ? Thanks. – 2017-01-05
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    I have $\omega = 2.24 /s$ – 2017-01-05
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    Yh I think because I rounded too early. – 2017-01-05
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    So the answer I got is a = 0.8 ms$^-2$ is this correct ? – 2017-01-05