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.

(1) The position around equilibrium of an object in simple harmonic motion is described by

where
A is the amplitude of the motion
is the angular frequency.

The velocity is the derivative of the position:

where 
is the maximum velocity of the object.

The acceleration is the derivative of the velocity:

where
is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:


From the first equation, we get
(1)
and if we substitute this into the second equation, we find the angular fequency

while the amplitude is  (using (1)):


(b) We found in the previous step that the angular frequency of the motion is

But the angular frequency is related to the period by

and so, the period is

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way