Simple Harmonic Motion help please! (1 Viewer)

[imsleepyz]

A particle is moving in a straight line with Simple Harmonic Motion. If the amplitude of the motion is 4cm and the period of the motion is 3 seconds, calculate:

i. the maximum velocity of the particle;
ii. the maximum acceleration of the particle;
iii. the speed of the particle when it is 2cm from the centre of the motion.

Any help is appreciated thank you

 

[Carrotsticks]

I assume that the centre of motion is at the origin?

Amplitude = 4

Period = 2pi/n = 3 ==> n = 3\2pi

So the equation is:



Differentiate this with respect to t to find velocity, and differentiate again to find acceleration as functions of time.

Maximum velocity occurs at the centre of motion (x=0) and maximum acceleration occurs at the endpoints of motion (x=plus minus 4).

To find the speed of the particle 2cm from centre of motion, sub x=plus/minus 2 into the velocity function.

 

[Meshy590]

why did u assume it's cos??

 

[Carrotsticks]

why did u assume it's cos??

Either one should still work. We only choose whether the function X is in terms of sine or cosine depending on the initial conditions.

ie: If it starts at the centre of motion, then sine is best. If it starts at an endpoint, cosine is best (and depending on which endpoint, we use positive or negative cosine).

The question didn't specify anything so I just used the cosine function.

 

[imsleepyz]

Thank you very much Carrotsticks!

Just one question, what exactly is the difference between maximum velocity and maximum acceleration? It seems the same...

 

[Carrotsticks]

Thank you very much Carrotsticks!

Just one question, what exactly is the difference between maximum velocity and maximum acceleration? It seems the same...

Many students get confused. Here's how I see it in my head but please bear with me because I often have 'strange' ways of understanding things.

When I think of Simple Harmonic Motion, I think of a Rollercoaster back and forth in a U shape.

Tell me, when do you think the most terrifying part of this rollercoaster would be? I would imagine it to be the very bottom of the U shape, and so that's where maximum VELOCITY occurs! Because that's when I'm going fastest.

Now, remember that acceleration is the change of VELOCITY. When do you think this would be the greatest? I would think it would be at the endpoints because that's the point where velocity goes from positive --> negative or visa versa! That's a pretty big change to me, and so that's where maximum acceleration occurs.

 

[imsleepyz]

Many students get confused. Here's how I see it in my head but please bear with me because I often have 'strange' ways of understanding things.

When I think of Simple Harmonic Motion, I think of a Rollercoaster back and forth in a U shape.

Tell me, when do you think the most terrifying part of this rollercoaster would be? I would imagine it to be the very bottom of the U shape, and so that's where maximum VELOCITY occurs! Because that's when I'm going fastest.

Now, remember that acceleration is the change of VELOCITY. When do you think this would be the greatest? I would think it would be at the endpoints because that's the point where velocity goes from positive --> negative or visa versa! That's a pretty big change to me, and so that's where maximum acceleration occurs.

That's a really well said analogy there Carrotsticks! Thanks heaps for that! I guess I can accept that maximum acceleration occurs at endpoints. Will this question of maximum acceleration ONLY be asked for simple harmonic motion?

 

[Carrotsticks]

Will this question of maximum acceleration ONLY be asked for simple harmonic motion?

Usually maximum acceleration is asked in SHM but it COULD be asked for other types of questions, but unlikely imo because the third derivative would be required in some cases, which is a lot surely!

 

[Sy123]

http://phet.colorado.edu/sims/pendulum-lab/pendulum-lab_en.html

Here is a great simulation displaying SHM, just imagine that the curved motion is stretched out on a straight line.
Just show the velocity and acceleration vector, let g=0.
At the end points, you notice that the velocity vector is not there, while acceleration is at its longest.

I just thought it would be great to share, because it helped me with the understanding of SHM

 

[Sanjeet]

I assume that the centre of motion is at the origin?

Amplitude = 4

Period = 2pi/n = 3 ==> n = 3\2pi

So the equation is:



Differentiate this with respect to t to find velocity, and differentiate again to find acceleration as functions of time.

Maximum velocity occurs at the centre of motion (x=0) and maximum acceleration occurs at the endpoints of motion (x=plus minus 4).

To find the speed of the particle 2cm from centre of motion, sub x=plus/minus 2 into the velocity function.

Why not
?

 

[Sy123]

We need to be given the displacement of x at a certain time to be able to evaluate that, but since the question has not stated there is no possible way to know what Alpha is in this.
Its not a good question considering Carrotsticks already had to assume that C.O.O is x=0

 

[Drongoski]

Why not
?

The question does not specify a particular SHM apart from the values of the key parameters. So it is OK and much easier to assume, say, particle starts from the originto x = +4 - in which case eqn is: - no need to complicate matters with the phase angle alpha.