John Harmonic Posted September 27, 2018 Share Posted September 27, 2018 Let us say you put the pendulum in position A (which requires energy), then you let it go and let it swing and swing until it suddenly stops. Would energy it took to put that pendulum in position A be equal to the amount of kinetic energy produced by the swing of the pendulum till it stops. Also what is the resonance of an oscillating pendulum. Link to comment Share on other sites More sharing options...
Strange Posted September 27, 2018 Share Posted September 27, 2018 3 hours ago, John Harmonic said: Let us say you put the pendulum in position A (which requires energy), then you let it go and let it swing and swing until it suddenly stops. It would not suddenly stop. It would slow down gradually due to air resistance and friction. The energy released (as heat) would equal the energy put in originally. 3 hours ago, John Harmonic said: Also what is the resonance of an oscillating pendulum. [math]T = 2 \pi \sqrt\frac{L}{g}[/math] (approximately) http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html Link to comment Share on other sites More sharing options...
DrP Posted September 27, 2018 Share Posted September 27, 2018 3 hours ago, John Harmonic said: Let us say you put the pendulum in position A (which requires energy), then you let it go and let it swing and swing until it suddenly stops. Would energy it took to put that pendulum in position A be equal to the amount of kinetic energy produced by the swing of the pendulum till it stops. The raising of it get it to position A gives the weight a potential energy. This potential energy is converted to KE as the pendulum falls.... the KE is converted back to potential energy again as the pendulum rises back to B and so on. It never quite gets back to point A or B after that due to some energy lost as friction. I would think that that the total energy lost in friction to heat and sound energy would equal the energy required to raise the thing to a starting position though. Thus - it would be more of a gradual stopping rather than a sudden stop. Link to comment Share on other sites More sharing options...
swansont Posted September 27, 2018 Share Posted September 27, 2018 If you mean stop at point B, then yes (for an ideal pendulum). The total energy when it's moving is always going to be KE + PE. At A and B the energy is all potential, and at the low point it's all kinetic. As others have mentioned, for a real pendulum there is always going to be air resistance and friction, which slowly reduces the total energy, which is why pendulum clocks have weights and gearing to give the pendulum a nudge to keep its amplitude constant. Link to comment Share on other sites More sharing options...
John Harmonic Posted September 28, 2018 Author Share Posted September 28, 2018 But it is weird, cause you only input a little bit of energy and then output is many swings till friction causes it to stop. Link to comment Share on other sites More sharing options...
Strange Posted September 28, 2018 Share Posted September 28, 2018 (edited) 6 hours ago, John Harmonic said: But it is weird, cause you only input a little bit of energy and then output is many swings till friction causes it to stop. It doesn’t lose much energy on each swing. Edited September 28, 2018 by Strange Link to comment Share on other sites More sharing options...
studiot Posted September 28, 2018 Share Posted September 28, 2018 (edited) Interestingly there is a current homework question about a pendulum receiving a small amout of energy. Question 69 here. Perhaps you should try it? You should not consider such phenomena in terms of energy input and output. They are not clearly defined terms in this case. As to resonance, a pendulum is the simplest device to understand the workings of the phenomenon, because it can be made to occur at human timescales. Most resonance phenomena are too fast for humans to follow directly. Edited September 28, 2018 by studiot Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now