Question

An oscillatory motion has a period of 50.0s and an amplitude of 20.0cm. Determine its maximum acceleration and mechanical energy

Answer 1

a)

$$a_{max}=-A\omega^2=-A\dfrac{4\pi^2}{T^2},$$ $$a_{max}=-0.2\ m\cdot\dfrac{4\pi^2}{(50.0\ s)^2}=-3.16\cdot10^{-3}\ \dfrac{m}{s^2}.$$

The magnitude of the maximum acceleration is $3.16\cdot10^{-3}\ \dfrac{m}{s^2}$.

b) By the definition of the law of conservation of energy, we have:

$$ME=U+K=\dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2.$$

An oscillatory motion of the oscillator is defined by the position:

$$x(t)=Acos(\omega t+\phi)$$

and velocity:

$$v(t)=-A\omega cos(\omega t+\phi).$$

Then, we can write:

$$ME=\dfrac{1}{2}kA^2cos^2(\omega t+\phi)+\dfrac{1}{2}mA^2sin^2(\omega t+\phi).$$

Applying trigonometric identity $cos^2\theta+sin^2\theta=1$ and $\omega=\sqrt{\dfrac{k}{m}}$, we get:

$$ME=\dfrac{1}{2}kA^2=\dfrac{1}{2}\cdot k\cdot(0.2\ m)^2=0.02\ m^2\cdot k.$$