Can velocity and acceleration reach maximal values during the SHM simultaneously? Can you explain why?

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    $\begingroup$ To be able to reach max velocity, the acceleration has to be zero, otherwise the velocity would still be changing (did not reach its max) or the acceleration would not be continues (jumping from something positive to something negative). $\endgroup$ – fibonatic Dec 16 '13 at 14:01

Consider a mass-spring system executing simple harmonic motion. If I draw the displacement, velocity and time graph, it would look something like this: enter image description here

You may see that when t=1 second, velocity is maximum and acceleration is zero. Another way to explain this is by using the definition of acceleration.

Acceleration = change in velocity/time = gradient of velocity time graph

In other words, when magnitude of velocity is maximum, there will be a stationary point. At that point, gradient is zero. Thus, acceleration is zero.

To answer your question, it is just not possible in SHM to have maximum acceleration and velocity. This is because if there is acceleration, velocity will continue to change (as acceleration is the rate of change of velocity)


As taught in calculus, to find the local extrema of a function, one solves for the values of the argument where the derivative of the function is zero.

So, at the maxima and minima of velocity, the (time) derivative of velocity is zero.

But acceleration is the (time) derivative of the velocity.

Do you see the answer now?


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