SHM question (1 Viewer)

[nazfiz]

A particle executes simple harmonic motion. When it is 2cm from its equilibrium position, its velocity is 6cm/s; when it is 3cm, its velocity is 4cm/s. Find the period of its motion and the amplitude.

Thanks! Any help will be repped.

 

[nazfiz]

bump.

 

[Alkenes]

use v^2= n^2 (a^2 - x^2)
sub in v n x ....get two equations and solve em' simultaneously

 

[nazfiz]

Oh ok, forgot about that! I'm new to this.
Thanks!

 

[Sy123]

Hang on, doesnt that formula only work if Centre of Oscillation is x=0?
Because thats where it is derived from.

It doesnt look like you can assume that x=0 is equilibrium either considering its careful wording (3cm from equilibrium), it would be pretty stupid if they didnt give you the equilibrium as zero, otherwise you would have to assume it is.

I did it the extremely hard way and was only able to get the period which by the way is pi. (I would latex it up, but it took me one page of working, so... yeah)

So yeah, Im not sure it is right to use that formula, because you cant just assume x=0 is equilibrium

 

[Sanjeet]

^Yup, given the information, you can only assume that the equilibrium is x=b (where b is any integer), so the equation of motion would be x = b +Asin(nt + α). Since you are only given two pieces of information, I'm not sure you can solve for the period and amplitude...I could be wrong though.

 

[Alkenes]

yeah sorry ... it is actually v^2= n^2 [a^2 - (x-b)^2]

 

[dulip]

Our teacher said we can't use that straight away, but must derive it from a = -n^2 x
is this true for HSC?

 

[Alkenes]

YES, you gotta derive it
it's not hard, is it?

Our teacher said we can't use that straight away, but must derive it from a = -n^2 x
is this true for HSC?

 

[Nooblet94]

^Yup, given the information, you can only assume that the equilibrium is x=b (where b is any integer), so the equation of motion would be x = b +Asin(nt + α). Since you are only given two pieces of information, I'm not sure you can solve for the period and amplitude...I could be wrong though.

You can define x=0 as being the equilibrium position. It doesn't say when x=3cm, v=4cm/s - instead it says 3cm from the equilibrium position, so you can define the equilibrium position as being any value of x.

Also, even if you could only assume it was at x=b there's no reason that b would have to be an integer.

 

[Carrotsticks]

You can define x=0 as being the equilibrium position. It doesn't say when x=3cm, v=4cm/s - instead it says 3cm from the equilibrium position, so you can define the equilibrium position as being any value of x.

Also, even if you could only assume it was at x=b there's no reason that b would have to b an integer.

This.

Also note that the equilibrium position does not matter because we are only concerned about the range in which the particle oscillates.

Suppose the equilibrium position were x=10000000, we could simply shift the system to be centred at x=0 to make calculations easier, because we see that doing so doesn't change anything.

 

[Sanjeet]

Ah I see, sorry.