Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Im trying to get my head around SMH out of curiosity because it seems simple yet I'm not getting the concept behind some ideas.

For a SMH equation :

$$ x=a \sin(\omega t+\phi) $$

  • Under what conditions will each of the two ( velocity and acceleration ) be at maximum ?

we know that :

$$ x'=-\omega \cos(\omega t + \phi) $$ $$ x''=-\omega^2 \sin(\omega t + \phi) $$

Also :

  • how would it be different for $x = a \cos(\omega t - \phi)$ ?
share|improve this question
    
I don't think this necessarily needs to have the homework tag. –  David Z Feb 19 '12 at 19:42
add comment

3 Answers

up vote 2 down vote accepted

The easiest way to determine maximum and minumums of a function is to set the derivative equal to zero. Thus, in this case, setting the equation for acceleration equal to zero and solving for the variables of interest will give you what you want. Thus, in this case we have the equation for position:

$$ x = a \sin(\omega t + \phi) $$

One way to see the maximum and minimum is to plot the function. I find that playing around with the constants (in this case $a, \omega,$ and $\phi$) helps you see what effect they have.

You can see that $a$ is an amplitude that determines how large the swings are from the origin. The termn $\omega$ tells you something about how quickly the position moves back and forth, thus it is known as the "angular frequency." Finally, $\phi$ tells you where you are when $t=0$ (or any other time for that matter) and is called the "phase."

For the position function, it is obvious that the value is maximized when the value of the sine function equals $1$ and, from what you learned in trig class, this happens when $\omega t + \phi = \pm \pi/2$ depending on the sign of $a$. Since $\omega$ and $\phi$ are constants set by the conditions and $t$ is the variable, the position is maximum when $t = (\pm \pi/2 - \phi)/\omega$, again depending on the sign of $a$.


ps- sorry for the curt answer the first time, but Wikipedia has a good article on this topic and there are tons of resources out there for it.

share|improve this answer
    
Its not really a homework question as I made it clear in the first sentence. I want to see how it works out differently because there are many websites online that just confused me. Usually people here answer with a way that I can directly relate to and thus understand the idea better. –  Fendi Feb 19 '12 at 16:34
    
Okay, did my answer work? or are you still confused? –  AdamRedwine Feb 19 '12 at 16:40
    
I still don't get it. Can you provide an example or explain more thoroughly please ? –  Fendi Feb 26 '12 at 18:10
    
If you give me more of a description of what you are thinking, it would be easier to give you a helpful answer. This might be a discussion better suited to the chat room. –  AdamRedwine Feb 27 '12 at 13:28
add comment

$$x'=-\omega \cos(\omega t + \phi)$$

maximum value of cosine is 1 implies max value of $x'$ is $|w|$.

Same for $x''$.

share|improve this answer
add comment

In simple harmonic motion displacement is directly proportional to acceleration.When the distance from the mean point(where the body has to stop)decreases the acceleration also decreases about velocity we all know that velocity is inversely proportional to distance from mean point mean as displacement increases the velocity decreases so when the body pass through the mean point the velocity is maximum and acceleration is 0. The most important thing to remember is that velocity is inversely proportional to displacement and acceleration is directly proportional to displacement. Here displacement means distance from mean point.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.