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Simple harmonic motion help Watch

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    Hello!

    Having trouble with a question i have to do in regards to simple harmonic motions topics.
    I seem to have worked out a) and b), but having trouble with part c).
    Im not sure how I would calculate this.

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    Here is my workings for part a and b. Would appreciate any feedback on this also.

    Attachment 615126615128

    Many thanks,
    Dan
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    What level is this? I don't want to confuse you with equations you won't have seen yet...
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    (Original post by Darth_Narwhale)
    What level is this? I don't want to confuse you with equations you won't have seen yet...
    Hello. its first term of HNC Engineering.

    Thanks.
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    Firstly your answer to B is wrong, the stiffness constant stays constant. What is happening is that someone has applied an extra force to displace the mass, and then removed that force so that it oscillates. Do you understand what is is you'v done wrong here?

    For (c), do you know the equation x = Acos(wt)? therefore, using differentiation, v = -Awsin(wt)
    This will be max when sin = 1
    Vmax = Aw

    Here, t is time, w is angular frequency (or 2*pi*frequency) = root(k/m), A is amplitude, x is displacement
    therefore Vmax = Aw = 20mm * root(245.25/2.25)
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    (Original post by Darth_Narwhale)
    Firstly your answer to B is wrong, the stiffness constant stays constant. What is happening is that someone has applied an extra force to displace the mass, and then removed that force so that it oscillates. Do you understand what is is you'v done wrong here?
    Ahh, so the K stiffness constant stays constant. got ya!

    So where do you add the extra force due to the extra extension length?
    I was working out the oscillation using:

    T = 2π √(M/K)
    T being period of oscillation.

    So I was doing T = 2π √(2.25/345.25 but I dont see how that takes into account any extra length/force?)

    (Original post by Darth_Narwhale)
    For (c), do you know the equation x = Acos(wt)? therefore, using differentiation, v = -Awsin(wt)
    This will be max when sin = 1
    Vmax = Aw

    Here, t is time, w is angular frequency (or 2*pi*frequency) = root(k/m), A is amplitude, x is displacement
    therefore Vmax = Aw = 20mm * root(245.25/2.25)
    Ive never really used Sin or Cos, but I will investigate that and have a go.

    Thanks,
    Dan
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    right, this might take a while to explain. I'm going to get you to derive a pretty key equation here.
    so we know that total force experience by the mass F = mg -kx (x is defined to be up here)

    The equilibrium, where the mass will happily sit without moving, is at F = 0 or mg = kx. We know this is when x = 90mm

    So now lets think about extra forces. I'm gonna use X to mean extension from the equilibrium point now.

    seeing as mg and k*90mm will always be present and will cancel out, lets just focus on the restorative force produced by any extra extension beyond 90mm.

    F = -kX (as extension down will produce a force upwards and visa versa)

    Now we also know that F = ma

    so ma = -kX, or x = -(k/m)a

    Have you done calculus yet? If so you'll recognise that this is a differential equation

    X = -(k/m) * (d/dt)(dX/dt)

    The solution to this equation is X = Acost(root(k/m)*t)
    or X = Acos(wt) where w = root(k/m)

    now you'll know, I'm sure, that cos(y) repeats itself every time y = 2π, so it's period is 2π
    cos(wt) repeats itself every time wt = 2π
    seeing as w is constant, t = 2π/w
    Handily, t is time, so we can just relable it T, because we are finding the specific time it takes for one oscillation, aka, the period.

    T = 2π/w
    w = root(k/m)
    Therefore T = 2πroot(m/k)

    You see that I arrived at this answer for T by first thinking about the extra force from the extra extension, so that is where it goes
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    There is another way to find the period of the spring, that does not involve using calculus.
    The basis is to recognise that if you were to take some particle moving in a circle, and then projected the shadow of that motion on a 1-d strip of paper, such that you get the horizontal component (or vertical. The point is that you get just one, which I am assuming to be horizontal for simplicity) you have the motion of the spring. I personally find this model very beautiful and amazing that nature has only one sort of periodic motion, but I digress. Basically the idea is that you can draw an analogy from circles.

    The period of a circle is T = 2πr/v where r is the radius and v the tangential velocity. Knowing that wr = v, where w is the angular speed gives the following: T = 2π/w. Now if you were to find the analogue for w in springs you get your period.

    To do this, use the centripetal acceleration: a = v^2/r = rw^2.
    Solving for w gives you:

     w = √\frac ar
    You need to find r and a in terms of springs.

    Hooke's law (with Newton's 2nd law) says that the acceleration of a spring is kx/m. Note that in our situation this acceleration is in fact the horizontal component of the acceleration an object would feel in a circle. In other words, a cos ø = kx/m, or solving for a:

     a = \frac{kx}{m cos ø}

    Now to find r. Note that the x displacement in the context of the spring is the horizontal component of r for circular motion. Thus, r cos ø = x, or solving for 1/r:

     \frac 1r = \frac {cos ø} {x}

    Therefore √a/r = w = cos ø * kx/m cos ø x , and cancelling the xs and cos øs gives:

     w = \sqrt {\frac km}

    Since T = 2π/w you need the reciprocal of w, which is simply √(m/k). Substituting this in T gives:

     T = 2\pi \sqrt {\frac mk}
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    As for the question, you can calculate it with energy conservation. The kinetic energy and thus the velocity is at a maximum when x = 0. The potential energy is at a maximum at the amplitude, 20 cm. By conservation of energy, these two must be equal, so 1/2mv^2 = 1/2kA^2 (Not sure if you have done this yet, if not see the note at the bottom). Solve for v.


    Note: The equation for potential energy from a spring can be determined by the work done against a spring. To pull a mass a distance x against the spring, you must extend the spring by x. In doing so, the spring exerts a force -kx. (The spring tries to pull the mass back, hence the sign). The potential energy is the negative of the work done by the spring as you pull against it, which is:

     \int kxdx  = 0.5kx^2

    The maximum potential energy is when v = 0, at the amplitude, where x = A..
 
 
 
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