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Gert
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You got stuck with the second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis), where $F$ is a constant force acting on the block:

$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$

But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.

Substitute:

$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$ So the differential equation is now homogeneous:

Set:

$$\frac km=\omega^2$$

$$\ddot{X}+\omega^2 X=0$$ This classic second order differential equation solves to:

$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:

Assume at $t=0$, $x=0 \implies X(0)=-F$

$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.

$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin\phi=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$x(t)=\frac Fk(1-\cos\omega t)$$$$\boxed{x(t)=\frac Fk(1-\cos\omega t)}$$ So this is the equation of motion. And this is the expression for velocity in time: $$\dot{x}=\frac Fk \omega \sin \omega t$$$$\boxed{\dot{x}=\frac Fk \omega \sin \omega t}$$

Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.

So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it. The object lesson is that if you're not tasked to find the exact trajectory of the object(s), solve this kind of problem with conservation, preferably.

You got stuck with the second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis), where $F$ is a constant force acting on the block:

$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$

But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.

Substitute:

$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$ So the differential equation is now homogeneous:

Set:

$$\frac km=\omega^2$$

$$\ddot{X}+\omega^2 X=0$$ This classic second order differential equation solves to:

$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:

Assume at $t=0$, $x=0 \implies X(0)=-F$

$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.

$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin\phi=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$x(t)=\frac Fk(1-\cos\omega t)$$ So this is the equation of motion. And this is the expression for velocity in time: $$\dot{x}=\frac Fk \omega \sin \omega t$$

Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.

So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it. The object lesson is that if you're not tasked to find the exact trajectory of the object(s), solve this kind of problem with conservation, preferably.

You got stuck with the second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis), where $F$ is a constant force acting on the block:

$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$

But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.

Substitute:

$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$ So the differential equation is now homogeneous:

Set:

$$\frac km=\omega^2$$

$$\ddot{X}+\omega^2 X=0$$ This classic second order differential equation solves to:

$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:

Assume at $t=0$, $x=0 \implies X(0)=-F$

$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.

$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin\phi=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$\boxed{x(t)=\frac Fk(1-\cos\omega t)}$$ So this is the equation of motion. And this is the expression for velocity in time: $$\boxed{\dot{x}=\frac Fk \omega \sin \omega t}$$

Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.

So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it. The object lesson is that if you're not tasked to find the exact trajectory of the object(s), solve this kind of problem with conservation, preferably.

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Gert
  • 35.5k
  • 8
  • 62
  • 107

You got stuck with the second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis), where $F$ is a constant force acting on the block:

$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$

But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.

Substitute:

$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$ So the differential equation is now homogeneous:

Set:

$$\frac km=\omega^2$$

$$\ddot{X}+\omega^2 X=0$$ This classic second order differential equation solves to:

$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:

Assume at $t=0$, $x=0 \implies X(0)=-F$

$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.

$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin(\phi)=0\implies \phi=0$$$$\dot{X}(0)=-A\omega\sin\phi=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$x(t)=\frac Fk(1-\cos\omega t)$$ So this is the equation of motion. And this is the expression for velocity in time: $$\dot{x}=\frac Fk \omega \sin \omega t$$

Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.

So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it. The object lesson is that if you're not tasked to find the exact trajectory of the object(s), solve this kind of problem with conservation, preferably.

You got stuck with the second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis), where $F$ is a constant force acting on the block:

$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$

But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.

Substitute:

$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$ So the differential equation is now homogeneous:

Set:

$$\frac km=\omega^2$$

$$\ddot{X}+\omega^2 X=0$$ This classic second order differential equation solves to:

$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:

Assume at $t=0$, $x=0 \implies X(0)=-F$

$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.

$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin(\phi)=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$x(t)=\frac Fk(1-\cos\omega t)$$ So this is the equation of motion. And this is the expression for velocity in time: $$\dot{x}=\frac Fk \omega \sin \omega t$$

Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.

So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it. The object lesson is that if you're not tasked to find the exact trajectory of the object(s), solve this kind of problem with conservation, preferably.

You got stuck with the second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis), where $F$ is a constant force acting on the block:

$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$

But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.

Substitute:

$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$ So the differential equation is now homogeneous:

Set:

$$\frac km=\omega^2$$

$$\ddot{X}+\omega^2 X=0$$ This classic second order differential equation solves to:

$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:

Assume at $t=0$, $x=0 \implies X(0)=-F$

$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.

$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin\phi=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$x(t)=\frac Fk(1-\cos\omega t)$$ So this is the equation of motion. And this is the expression for velocity in time: $$\dot{x}=\frac Fk \omega \sin \omega t$$

Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.

So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it. The object lesson is that if you're not tasked to find the exact trajectory of the object(s), solve this kind of problem with conservation, preferably.

added 51 characters in body
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Gert
  • 35.5k
  • 8
  • 62
  • 107

You got stuck with the second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis), where $F$ is a constant force acting on the block:

$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$

But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.

Substitute:

$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$ So the differential equation is now homogeneous:

Set:

$$\frac km=\omega^2$$

$$\ddot{X}+\omega^2 X=0$$ This classic differential equationsecond order differential equation solves to:

$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:

Assume at $t=0$, $x=0 \implies X(0)=-F$

$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.

$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin(\phi)=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$x(t)=\frac Fk(1-\cos\omega t)$$ So this is the equation of motion. And this is the expression for velocity in time: $$\dot{x}=\frac Fk \omega \sin \omega t$$

Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.

So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it. The object lesson is that if you're not tasked to find the exact trajectory of the object(s), solve this kind of problem with conservation, preferably.

You got stuck with second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis):

$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$

But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.

Substitute:

$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$

Set:

$$\frac km=\omega^2$$

$$\ddot{X}+\omega^2 X=0$$ This classic differential equation solves to:

$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:

Assume at $t=0$, $x=0 \implies X(0)=-F$

$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.

$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin(\phi)=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$x(t)=\frac Fk(1-\cos\omega t)$$ So this is the equation of motion. And this is the expression for velocity in time: $$\dot{x}=\frac Fk \omega \sin \omega t$$

Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.

So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it. The object lesson is that if you're not tasked to find the exact trajectory of the object(s), solve this kind of problem with conservation, preferably.

You got stuck with the second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis), where $F$ is a constant force acting on the block:

$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$

But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.

Substitute:

$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$ So the differential equation is now homogeneous:

Set:

$$\frac km=\omega^2$$

$$\ddot{X}+\omega^2 X=0$$ This classic second order differential equation solves to:

$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:

Assume at $t=0$, $x=0 \implies X(0)=-F$

$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.

$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin(\phi)=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$x(t)=\frac Fk(1-\cos\omega t)$$ So this is the equation of motion. And this is the expression for velocity in time: $$\dot{x}=\frac Fk \omega \sin \omega t$$

Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.

So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it. The object lesson is that if you're not tasked to find the exact trajectory of the object(s), solve this kind of problem with conservation, preferably.

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Gert
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Gert
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Gert
  • 35.5k
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  • 62
  • 107
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