Helped needed for oscillation concepts?

Question

For oscillations: omega^2 x amplitude = acceleration, i thought it should be acceleration=-omega^2xdisplace...
Is displacement always equal to amplitude?
Where is the negative sign and when do we have to take into consideration the negative sign?

I totally cannot understand how the equations are used in the topic oscillations are used and i always fail with 0 marks cause i don't even know what formula to use, can someone please explain to me.

like i saw v=+/- (omega x square root of x0^2-x^2) what is x0?
and why i see another formula for velocity to be -x0sin(omega x t)
where t=time???

what is the difference between x0 and amplitude, i thought they are the same?

HELP, i can't understand and i am suffering from a mental breakdown at the moment.

Please reply fast!!!

Answer 1

"For oscillations: omega^2 x amplitude = acceleration, i thought it should be acceleration=-omega^2xdisplace"

better put as "magnitude of maximum acceleration = omega^2 x amplitude" - this expression only applies to body at the amplitude position

"Is displacement always equal to amplitude?"
no - amplitude its the maximum value of displacement - when the object is furthest from equilibrium (the middle)"

"Where is the negative sign and when do we have to take into consideration the negative sign?"

acceleration = - omega^2 x displacement, so acceleration is positive when displacement is negative (and vice versa). You need a co-ordinate system e.g up is positive, down is negative or right is positive, left is negative, like a graph



"like i saw v=+/- (omega x square root of x0^2-x^2) what is x0?"
x0 here is amplitude

"and why i see another formula for velocity to be -x0sin(omega x t)
where t=time???"

Basically 2 formulae - the first tells you the velocity at different positions (displacements) in a cycle; it tells you that the velocity is fastest in the middle, zero at the edges; the second tells you at different times, so it follows a sine wave to show how the velocity rises and falls

"what is the difference between x0 and amplitude, i thought they are the same?" Yes

Answer 2

Look, it's easy if you go step by step.

Amplitude is the maximum value of the displacement.

The displacement is in general different of the amplitude and it's described by

x(t) = A cos (omega * t)

where x is the displacement, A is the amplitude omega is the angular frequency and t is the time. If you take values of time such that cos (omega * t) is 1, x is there equal to A. Just in those times the displacement equals the Amplitude. Where cos (omega * t) is -1 you will have x = -A and there you say that the displacement equals in absolute value the Amplitude.

If you take the first derivative of x(t) with respect to time, you will obtain the velocity

v(t) = - A*omega*sin(omega * t)

and then if you take the next derivative with respect to time you will obtain the acceleration

a(t) = - A omega^2 cos(omega * t)

you can rearrange this as

a(t) = - omega^2 {A cos(omega * t)}

the expression in {} is the displacement and then you have

a(t) = - omega^2 x(t)

Now, x0 here represents your Amplitude, there's no problem, you can name it as you want and instead of naming A you can name it x0. If we write now for x0 as the amplitude

x(t) = x0 cos(omega * t)

the velocity is

v(t) = - x0 omega sin(omega * t)

you can rewrite as

v(t) = - omega {x0 sin(omega * t)}

take the 2nd power and the sq root of the expression in { }

v(t) = - omega sqrt{x0^2 sin^2(omega * t)}

we know that sin^2(omega * t) = 1 - cos^2(omega * t), then

v(t) = - omega sqrt{x0^2 [1 - cos^2(omega * t)] }

just distribute

v(t) = - omega sqrt{x0^2 - x0^2 cos^2(omega * t)}

and the second term is nothing but {x(t)}^2 or just x^2

v(t) = - omega sqrt{x0^2 - x^2}

here is your equation.

Answer 3

For oscillations

Displacement varies with time.

The maximum displacement is called amplitude.

From your question,

The relation between displacement and time is given by the equation

x (t) = x (0) cos w*t. I have used “w” for omega.

x(0) is the maximum displacement ( amplitude) and this is when t = 0.

When t = 0, cos wt = 1 and hence x(t) = x(0)

When wt = pi/ 2, cos pi/2 = 0 and hence x(t) = 0
The displacement is zero.

When wt = pi, cos pi = -1 and hence x(t) = -x(0)
The displacement is the maximum in the opposite direction.

When wt = 3pi/2, cos 3pi/2 = 0 and hence x(t) = 0
The displacement is zero.

When wt = 2pi, cos 2pi = 1 and hence x(t) = x(0)
The displacement is the maximum and is equal to the initial displacement.

Thus at the start the body in oscillation has displaced to its maximum value x(0) or amplitude.

x(t) gives the displacement at any time t.

The relation for velocity and time is
v(t) = -v(0) sin w*t, where v(0) = w*x(0)

You have stated wrongly as x(0) sin wt.

In time t = 0, sin wt = 0, hence v (t) = 0.

As before, you can find the velocity at any time t.

The velocity is maximum when sin wt is the maximum and it is equal to one when wt = pi/2.

The maximum velocity is v(t) maximum = v(0) = w x(0)

This means at the maximum displacement or at amplitude, its velocity is zero.

The negative sign indicates that the direction of displacement and the velocity are opposite to each other.

Similarly you can find for acceleration.

The relation of acceleration with time is

a(t) = -a(0) cos wt, where a(0) = w^2*x(0)

The maximum value of acceleration is when cos wt = 1,

and the value is w^2*x(0).

Answer 4

I'm sorry, but my Physics course does not cover oscillations. I will tell you one thing though, displacement is NOT always equal to amplitude. It is only equal to amplitude when you are pulling from a completely non-tensioned spring. If you are pulling from somewhere where the spring is already exerting an elastic force, then the displacement will not equal the amplitude. Good luck with your oscillations, and sorry i could not be of more help.