- #1
fara0815
- 45
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Hello there!
I started a new chapter in my script and have done some problems about harmonic motion. After searching in this forum for a similar problem for ages but without any succees, I need to ask you guys to help me out.
Here it is:
A toy car of mass 1kg is attached with a spring to a wall. The spring's constant is 20 N/m. The car is beeing pulled 1 m (horizontal) away from the wall and then let go. It is going back towards the wall up to a distance of 0.4 m.
What is the system's energy?Every way I took led me into a death end. The most logical way to me is to find first the equilibrium point of the spring which is supposed to be at the same spot where the velocity is the greatest:
[tex] V_{max}= A\times \omega [/tex]
I use 0.6 m for A and I compute omega with
[tex]\omega=\sqrt{\frac{k}{m}}= 0.6m\sqrt{20}= 2.68 m/s[/tex]
The equilibrium point is where KE and KP are equal:
[tex] \frac{1}{2}mv^2=\frac{1}{2}kx^2[/tex]
that gives me the x= 0.366 m.
My guess is that the equilibrium point is 1m - 0.366m.
From here I do not have any clue how to get the over all energy of the mass-spring system.
I would be very happ about any kind of help !
I started a new chapter in my script and have done some problems about harmonic motion. After searching in this forum for a similar problem for ages but without any succees, I need to ask you guys to help me out.
Here it is:
A toy car of mass 1kg is attached with a spring to a wall. The spring's constant is 20 N/m. The car is beeing pulled 1 m (horizontal) away from the wall and then let go. It is going back towards the wall up to a distance of 0.4 m.
What is the system's energy?Every way I took led me into a death end. The most logical way to me is to find first the equilibrium point of the spring which is supposed to be at the same spot where the velocity is the greatest:
[tex] V_{max}= A\times \omega [/tex]
I use 0.6 m for A and I compute omega with
[tex]\omega=\sqrt{\frac{k}{m}}= 0.6m\sqrt{20}= 2.68 m/s[/tex]
The equilibrium point is where KE and KP are equal:
[tex] \frac{1}{2}mv^2=\frac{1}{2}kx^2[/tex]
that gives me the x= 0.366 m.
My guess is that the equilibrium point is 1m - 0.366m.
From here I do not have any clue how to get the over all energy of the mass-spring system.
I would be very happ about any kind of help !
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