Maximum displacement from equilibrium of the mass as it osciallates

[Reema]

A mass of 0.1 kg connected to a spring with a spring constant 10 N/M oscillates horizontally on frictionless table. The speed of the mass is 3.0 m/s when the displacement is 0.2 m form its equilibrium position.

What is the maximum displacement from equilibrium of the mass as it oscillates?

So ... the answer is 0.361m, but I did it one hundered times and I always get 0.969


First I found the frequency

I got 1.59 HZ so w= 9.99

then I used V= w ( √ A^2 - X^2 )



A^2= v^2/w + X^2

= 3^2 / 9.99 + o.2^2

A^2 = 0.94

A= 0.969m

 

[Leveret]

Your method seems right to me. Check your algebra, though.

 

[Reema]

Thanks Leveret

do you think that there is something wrong with this equation A^2=( v^2/w) + X^2
because I thought it might be A^2=( v^2/w^2) + X^2 ?

 

[lackos]

Try looking at it from an energy conservation perspective.

are you aware of that formula?

its a much simpler method

 

[Reema]

dou you mean (PE = KE )

 

[lackos]

actually for simple harmonic motion (oscillation) there is a particular formula.
the formula is:

(1/2)kA^2 = (1/2)mv^2 + (1/2)kx^2

where (1/2)kA^2 is the total enegy, (1/2)kx^2 is the potential energy and (1/2)mv^2 is the kinetic energy

rearrange and you should be sweet :)also your method is correct ( in fact your equations are derived from the one i gave you). your mistake was you didnt square the omega

 

[Reema]

Wow .. Finally I got 0.36 :D

thaaaaaanks alot