Question

An object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s2 . find (a) the amplitude and (b) the period of this motion.

Answer 1

(1) The position around equilibrium of an object in simple harmonic motion is described by
$x(t) = A \cos (\omega t)$
where
A is the amplitude of the motion
$\omega$ is the angular frequency.

The velocity is the derivative of the position:
$v(t)=-\omega A \sin(\omega t) = -v_0 \sin (\omega t)$
where 
$v_0 = \omega A$ is the maximum velocity of the object.

The acceleration is the derivative of the velocity:
$a(t)=- \omega^2 A \cos (\omega t) = -a_0 \cos (\omega t)$
where
$a_0=\omega^2 A$ is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:
$v_0 = \omega A = 4.3 m/s$
$a_0 = \omega^2 A = 0.65 m/s^2$
From the first equation, we get
$A= \frac{4.3 }{\omega}$ (1)
and if we substitute this into the second equation, we find the angular fequency
$\omega=0.15 rad/s$
while the amplitude is  (using (1)):
$A=28.7 m$


(b) We found in the previous step that the angular frequency of the motion is
$\omega=0.15 rad/s$
But the angular frequency is related to the period by
$T= \frac{2 \pi}{\omega}$
and so, the period is
$T= \frac{2 \pi}{\omega}= \frac{2 \pi}{0.15 rad/s}=41.9 s$