SHM - Sign Taking (1 Viewer)

[Lukybear]

For SHM say this Q
A particle executes SHM with n =pi/4 a =10

Find v and aceleration at x=6

When finding v

v^2 = (pi/4)^2(64)

How do we take the sign from here on?


Also for max acceleration i argued that it occurs at x=0
but acceleration then = -3pi^2/8

So what reasoning is used to say that acceleration is -3pi^2/8

 

[Lukybear]

Also i have 1 more question

Given at t=0 x=0, t=11 x=0.5, t=18 x=0

Find x at t=7, 9

Assuming particle moves in straight line with
a) unifrom acceleration
b) SHM with period of 12 seconds

 

[Aquawhite]

For SHM say this Q
A particle executes SHM with n =pi/4 a =10

Find v and aceleration at x=6

When finding v

v^2 = (pi/4)^2(64)

How do we take the sign from here on?


Also for max acceleration i argued that it occurs at x=0
but acceleration then = -3pi^2/8

So what reasoning is used to say that acceleration is -3pi^2/8

In regard to what sign you take... the question will usually guide you. Generally you can use both, but you may find that taking the negative velocity may not work out in the end. You can simply take velocity as a non-vector also, except you should specify this if you want to just work with one answer.

Also, in regard to acceleration... Acceleration is maximum when the velocity is equal to zero: you can work this out using your equations.

The reasoning behind a = max when v = 0:

Think of a normal pendulum, the max velocity will occur in the middle of the period; and then stop at either end. When the bob of the pendulum has stopped, there is a huge 'potential energy' (hope this isn't too Physics) and then you have a maximum acceleration for the particle.

You just have to remember that.

 

[cutemouse]

Generally you can use both

This is wrong. You need to take the sign that the initial conditions given. ie, if velocity was initially in a positive direction, then you take the +ve case.

Acceleration is maximum when the velocity is equal to zero: you can work this out using your equations.

This is also wrong. Max acceleration occurs at the end points of motion.

 

[Aquawhite]

This is wrong. You need to take the sign that the initial conditions given. ie, if velocity was initially in a positive direction, then you take the +ve case.


This is also wrong. Max acceleration occurs at the end points of motion.

In regard to taking the sign in accordance to the initial conditions... I did hint at that "the question will usually guide you"; however, the question will not always say if it travels in a certain direction or velocity and may be part of the question to figure that out.

"Max acceleration occurs at the end points of motion". Indeed it does... and you will also find that this is where velocity is equal to zero, since in SHM, the particle must swing back into the harmonic motion. How are you meant to sub in 'end points of motion' into an equation to find the maximum velocity?

 

[fullonoob]

speed = |"velocity"| -> the sign of v indicates the direction of motion
acceleration = " " + or - (travelling left or right)
max acceleration when v=0, not moving

 

[Lukybear]

What about for this particular case?

When finding v

v^2 = (pi/4)^2(64)

How do we take the sign from here on?

 

[mirakon]

This is also wrong. Max acceleration occurs at the end points of motion.

Um, v=0 at end points of motion? So he is in fact right and you are wrong.

Also in terms of LUkyBear's question it's positive velocity

 

[hscishard]

speed = |"velocity"| -> the sign of v indicates the direction of motion

I don't think that's right in physics...

 

[Trebla]

Speed = | Velocity | is definitely true for instantaneous situations, however once you start talking average speed and average velocity (i.e. over discrete non-infinitesimal time) which is often done in high school physics this equality breaks down.

 

[ohexploitable]

I don't think that's right in physics...

Yar it is, like -10 ms^-1 North = 10 ms^-1.

 

[hscishard]

Speed = | Velocity | is definitely true for instantaneous situations, however once you start talking average speed and average velocity (i.e. over discrete non-infinitesimal time) which is often done in high school physics this equality breaks down.

I see

Yar it is, like -10 ms^-1 North = 10 ms^-1.

I was thinking distance/time and displacement/time.

 

[ohexploitable]

however once you start talking average speed and average velocity (i.e. over discrete non-infinitesimal time) which is often done in high school physics this equality breaks down.

But this.

 

[hscishard]

But this.

Yea I know now.
Haven't done motion yet.

 

[PilotMerp]

Um, v=0 at end points of motion? So he is in fact right and you are wrong.

Actually since they are both right (velocity is zero at the end points with max accel)

YOU are in fact incorrect in stating that one of them is in correct/incorrect

SINCE they are both Correct