Physics Help Forum Amplitude

 Equilibrium and Elasticity Equilibrium and Elasticity Physics Help Forum

 Apr 19th 2011, 10:22 AM #1 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Amplitude It's only now that I realised that I wasn't quite sure how the amplitude in a simple pendulum is worked out. If we have a simple pendulum in motion for instance, what is the amplitude (in metres)? I know that for a body attached to a spring and oscillating on a frictionless surface, the amplitude is given by x - x_o where x is the maximum horizontal displacement and x_o is the equilibrium position of the body. When it comes to a pendulum, can we take the amplitude as the horizontal displacement from the equilibrium position or we take into consideration the vertical and horizontal displacement together (and ultimately using Pythagoras' Theorem) Say (fictive figures) that the maximum height the bob of a pendulum gains is 0.01 m relative to its initial vertical position and the maximum horizontal distance from the equilibrium position is 5 cm. Strictly speaking, is amplitude \sqrt(0.01^2 + 0.05^2) ? For instance, is it the value I use in, for example a = -ω^2x (a = maximum acceleration, ω = angular velocity, x = maximum displacement) __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
Apr 19th 2011, 11:10 AM   #2

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 Originally Posted by Unknown008 It's only now that I realised that I wasn't quite sure how the amplitude in a simple pendulum is worked out. If we have a simple pendulum in motion for instance, what is the amplitude (in metres)? I know that for a body attached to a spring and oscillating on a frictionless surface, the amplitude is given by x - x_o where x is the maximum horizontal displacement and x_o is the equilibrium position of the body. When it comes to a pendulum, can we take the amplitude as the horizontal displacement from the equilibrium position or we take into consideration the vertical and horizontal displacement together (and ultimately using Pythagoras' Theorem) Say (fictive figures) that the maximum height the bob of a pendulum gains is 0.01 m relative to its initial vertical position and the maximum horizontal distance from the equilibrium position is 5 cm. Strictly speaking, is amplitude \sqrt(0.01^2 + 0.05^2) ? For instance, is it the value I use in, for example a = -ω^2x (a = maximum acceleration, ω = angular velocity, x = maximum displacement)
Hmph! I have never thought of this question. I've never seen the SHO equation put to use on a pendulum, just talk about the pendulum's period. On the other hand we can write a SHO equation to describe the angle the pendulum is at. That is to say
(theta)(t) = (theta0)*cos(wt + phi)
where w = sqrt{g/L}

This is the solution to the (approximate) equation
(theta)'' + (g/L)(theta) = 0

But again, I don't think I've ever seen anyone talk about the general solution, just the period.

-Dan
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 Apr 19th 2011, 11:33 AM #3 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Indeed, I too have realised this and I started wondering and searching on the web didn't bring up a satisfactory answer for me. The sites I got were vague and not explicit enough. Wait, now I'm thinking about something... Okay, let's say the bob is raised (keeping the thread of the pendulum taut) through 1 m and the length of the string is 100 m (for ease of working). PE = mgh = 1*10*1 = 10 J (taking m = 1 kg XD and g = 10 m/s^2 again, for calculations purposes) Max KE = 10 J so, max v = \sqrt(10*2/1) = 2\sqrt(5) We can also find omega. ω = \sqrt{g/l} = \sqrt{10/100} = \sqrt{10}/10 we know that v = ω \sqrt{x_o^2 - x^2} 2\sqrt{5} = \sqrt{10}/10 x_o squaring, 20/ \sqrt{2} = x_o Now, from trigonometry (assuming that amplitude is displacement between the two points), the amplitude would be given by: Amplitude = \sqrt{x^2 + y^2} = \sqrt{ (\sqrt{100^2-99^2})^2 + 1^2} (A sketch shows it all) So, displacement = \sqrt{199 + 1} = \sqrt{200} = 10\sqrt{2} *looks up* Hm... 20/ \sqrt{2} = 20\sqrt{2} / 2 = 10\sqrt{2} Okay, now I'm satisfied. I'm not sure why I didn't think of this earlier >.< Maybe it's your answer which triggered it... __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?

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