SHO stuck on a simple question -.-

  • #1

Homework Statement



A horizontal massless spring of spring constant k is attached to a immovable wall at one end and a mass of 0.45kg at the other end. The spring which was not originally under tension, is now extended by 0.18 m by pulling the mass horizontally. The mass was then released by giving it an instantantaneous push in the direction of the extension such that it had an initial velocity of 0.5 ms-1. The mass then undergoes firctionless simple harmonic motion (SHM) at [tex]\omega[/tex] = 2.2 rad s-1. With an explanation of your reasoning:

(i) deduce the equation for the displacement of the mass as a function of time, (2)

(ii) show that your equation complies with the general form of SHM

x(t) = Acos([tex]\omega[/tex]ot + [tex]\phi[/tex]) (2)
 

Answers and Replies

  • #2
woah why did this get moved? lol this is 2nd year physics :p and i saw a fellow forum user ask about a 1st year physics question in the advance section. no fair ;[
 
  • #3
rock.freak667
Homework Helper
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31
For SHM you can either have x=x0sinwt or x=x0coswt

So why did you pick [itex]x=x_0cos\omega t[/itex] instead of [itex]x=x_0sin\omega t[/itex] ?
 
  • #4
For SHM you can either have x=x0sinwt or x=x0coswt

So why did you pick [itex]x=x_0cos\omega t[/itex] instead of [itex]x=x_0sin\omega t[/itex] ?
I did not pick it that is the second part of the question -.-
 
  • #5
rock.freak667
Homework Helper
6,230
31
I did not pick it that is the second part of the question -.-
then what exactly did you try in this question? (I ask because you did not post any attempt)
 
  • #6
this is not a homework question it is a past exam question and i dont know where to begin. :S
 
  • #7
rock.freak667
Homework Helper
6,230
31
this is not a homework question it is a past exam question and i dont know where to begin. :S
Well what is the force acting on the spring with extension x? When you get that recall that F=ma and a=d2x/dt2
 
  • #8
Well what is the force acting on the spring with extension x? When you get that recall that F=ma and a=d2x/dt2
force acting on a spring is f = -kx and [tex]\ddot{}x[/tex] = [tex]\omega[/tex]o2x

i know how to arrive to that equation f = - kx but also f = ma therfore ma = -kx re arrange and you get [tex]\ddot{}x[/tex] = [tex]\omega[/tex]o2x
 
  • #9
ideasrule
Homework Helper
2,266
0
If ma=-kx, then ma+kx=0 and
[tex]
\ddot{}x+\omega^2x=0
[/tex]

So you were off by a negative sign.

Now show that this equation satisfies this condition:

x(t) = Acos([tex]\omega[/tex]ot + [tex]\phi[/tex])

In other words, show that the second time derivative of x(t) is indeed equal to [tex]
-\omega^2x
[/tex]

That would show x(t) = Acos([tex]\omega[/tex]ot + [tex]\phi[/tex]) is a solution to the differential equation and therefore describes simple harmonic motion.
 
  • #10
If ma=-kx, then ma+kx=0 and
[tex]
\ddot{}x+\omega^2x=0
[/tex]

So you were off by a negative sign.

Now show that this equation satisfies this condition:

x(t) = Acos([tex]\omega[/tex]ot + [tex]\phi[/tex])

In other words, show that the second time derivative of x(t) is indeed equal to [tex]
-\omega^2x
[/tex]

That would show x(t) = Acos([tex]\omega[/tex]ot + [tex]\phi[/tex]) is a solution to the differential equation and therefore describes simple harmonic motion.


the original question asked for an equation as a function of time,
[tex]

\ddot{}x+\omega^2x=0

[/tex]

there is no time in that equation o_O
 
  • #11
954
117
the original question asked for an equation as a function of time,
[tex]

\ddot{}x+\omega^2x=0

[/tex]

there is no time in that equation o_O
Yes there is...[tex]x\ddot{}[/tex] is the second derivative of x wrt time
 
  • #12
Yes there is...[tex]x\ddot{}[/tex] is the second derivative of x wrt time
oh yeah there is
 

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