How come you have to use work energy to find v0 of spring?

[annamal]

TL;DR Summary

If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity if you pulled the spring a distance x. How come you cannot do kx*(delta t) = m*v to get the initial velocity and what would be the delta t value?

If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity if you pulled the spring a distance x. How come you cannot do kx*(delta t) = m*v to get the initial velocity and what would be the delta t value?

 

[PeroK]

TL;DR Summary: If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity if you pulled the spring a distance x. How come you cannot do kx*(delta t) = m*v and what would be the delta t value?

If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity if you pulled the spring a distance x. How come you cannot do kx*(delta t) = m*v and what would be the delta t value?

Because in the first case you have the energy at a point $x_1$ (it's better to use this to see that it's a fixed point, not a variable).

In the second case, $x$ is variable and the applied force varies with time.

 

[annamal]

Because in the first case you have the energy at a point $x_1$ (it's better to use this to see that it's a fixed point, not a variable).

In the second case, $x$ is variable and the applied force varies with time.

I am just looking for the initial velocity though

 

[PeroK]

I am just looking for the initial velocity though

The initial velocity is zero, I assume.

 

[annamal]

The initial velocity is zero, I assume.

I mean what is the initial velocity once you let go of the spring pulled a certain distance x. Why can't you do that with kx*(delta t) and what is delta t

 

[Ibix]

If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity

This is not correct, which may be the source of your confusion. A correct statement would be that $\frac 12kx^2+\frac 12mv^2=\mathrm{const}$, which you can also get by considering forces if you want.

If you solve for the $\mathrm{const}$ you will find that it is $\frac 12kx_{\mathrm{max}}^2$, where $x_{\mathrm{max}}$ is the maximum value of $x$. If you then consider the case where $x=0$ you will find that this is where velocity is a maximum, which leads to $\frac 12mv_{\mathrm{max}}^2=\frac 12kx_{\mathrm{max}}^2$, which may be what you are thinking of, but only determines the speed at equilibrium, not the initial speed.

 

[PeroK]

I mean what is the initial velocity once you let go of the spring pulled a certain distance x.

The initial velocity is zero. The mass then accelerates over time, under a changing force.

Why can't you do that with kx*(delta t) and what is delta t

For small $\Delta t$, we have $F \approx kx_{max}$ and $mv \approx F\Delta t$. That only works for a short time, before you have to take the changing force into account.

 

[sophiecentaur]

I mean what is the initial velocity once you let go of the spring pulled a certain distance x.

The initial velocity is whatever it's set to by the experimenter (perhaps zero). Your calculation with kx will tell you the Initial Acceleration of the mass and not the velocity

 

[sophiecentaur]

TL;DR Summary: If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity if you pulled the spring a distance x. How come you cannot do kx*(delta t) = m*v to get the initial velocity and what would be the delta t value?

How come you cannot do kx*(delta t) = m*v to get the initial velocity and what would be the delta t value?

The fact that you need to ask what delta t would be is the answer to 'why'. You want to use momentum instead of Energy so there must be a method to do it that way but using Potential Energy transfer to Kinetic Energy avoids all that hassle. If you do all that correctly, you would get the same (right) answer but life's too short.

The good thing about the Energy way is that Energy is conserved throughout the process.

 

[vanhees71]

Or to put it differently: The work-energy theorem holds for any force, i.e., from the equation of motion
$$m \dot{\vec{v}}=\vec{F},$$
where $\vec{F}$ can be an arbitrary function of $\vec{x}$, $t$, $\dot{\vec{x}}$(, and even higher time derivatives of $\vec{x}$, but this usually leads to other trouble). Now multiply this with $\vec{v}$ (using the scalar product). Then you get
$$m \vec{v} \cdot \dot{\vec{v}}=\vec{v} \cdot \vec{F}.$$
Now the left-hand side is a total time derivative, i.e.,
$$m \vec{v} \cdot \dot{\vec{v}}=\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \vec{v}^2 \right)=\dot{E}_{\text{kin}}=\vec{v} \cdot \vec{F}.$$
Now integrate this wrt. time over a time interval $(t_1,t_2)$ this gives the work-energy theorem,
$$E_{\text{kin}}(t_2)-E_{\text{kin}}(t_1)=\int_{t_1}^{t_2} \mathrm{d} t \vec{v} \cdot \vec{F}=W(t_1,t_2),$$
i.e., the change in kinetic energy is given by the work done in the said time interval.

This doesn't help much, if you don't know the solution of the equation of motion, and it's thus a pretty dull result. This drastically changes, if $\vec{F}$ is a potential field, i.e., if it's a function of $\vec{x}$ only, and it can be written as a gradient of a scalar field $V$. Usually one defines
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
Using this in the work-energy theorem leads to
$$W(t_1,t_2)=-\int_{t_1}^{t_2} \mathrm{d} t \dot{\vec{x}} \cdot \vec{\nabla} V(\vec{x}) = -\int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} V(\vec{x})=-V(\vec{x}_2)+V(\vec{x}_1),$$
where $\vec{x}_2=\vec{x}(t_2)$ and $\vec{x}_1=\vec{x}(t_1)$. So you have from the work energy theorem
$$E_1=E_{\text{kin} 1}+V(\vec{x}_1)=E_2=E_{\text{kin} 2} +V(\vec{x}_2),$$
i.e., the then definable total energy,
$$E=E_{\text{kin}}+V(\vec{x})=\text{const},$$
is conserved for any solution of the equation of motion, i.e., you don't need to know the specific equation of motion, to have some information from enery conservation.

For a force in 1D motion, $F(x)=-k x$ you have $V(x)=k x^2/2$ and thus
$$\frac{m}{2} v^2 + \frac{k}{2} x^2=\text{const}.$$
I'm not sure about the initial conditions in the OP. I guess, at $t=0$ you expanded the spring by $x_0>0$ from the equilibrium position $x=0$ and then realease it, i.e., your initial conditions are $x(0)=x_0$ and $v(0)=0$. Then the energy-conservation law tells you that
$$E=\frac{m}{2} v^2 + \frac{k}{2} x^2=E_0=\frac{k}{2} x_0^2.$$
This tells you, without solving for the equations of motion that you always must have $|x|<x_0$ and that the maximal velocity is for $x=0$, and that this maximal speed is given by
$$\frac{m}{2} v_{\text{max}}^2=E_0=\frac{k}{2} x_0^2 \; \Rightarrow \; |v_{\text{max}}|=\sqrt{\frac{k}{m}} x_0.$$
For $|x|=x_0$ you always have $v=0$. This tells you that the mass oscillates between the points $x=x_0$ and $x=-x_0$, and at this turning points $\dot{x}=v=0$. So you get a pretty good qualitative picture about the motion, without solving for the equation of motion, just from the energy-conservation law, which holds for the forces, which have a potential in the above described sense.