Homework Help Forum: Physics

Posted by Ellie on Wednesday, November 4, 2009 at 11:05pm.

You want to make an old-fashioned pendulum clock, with a period of exactly 1.4 second(s). Use the Earth's gravitational acceleration g = 9.8 m/s2. a) Ignoring the size of the pendulum bob, from what length wire should you hang your pendulum bob? .4865 m b) If you release the bob from an angle of 3.0°, what is the approximate acceleration of the bob at release? (you can use the approximation, sin(θ) = θ) m/s2 c) Describing a pendulum as a harmonic oscillator requires the use of the small angle approximation. What is the magnitude of the difference between the actual acceleration (without using the small angle approximation) and the approximate acceleration (using the small angle approximation)? The answer is a very small number. As for all other numerical answers, provide your answer with 4 significant digits. m/s2 d) From this release angle, what is the change in the bob's height from release to the bottom of its arc? m e) What is the bob's speed at the bottom of its arc? (use conservation of energy) m/s answered first not sure what formula to use on the rest can someone help?

• Physics - Ellie, Thursday, November 5, 2009 at 12:07am

  b i got a=gsin(3)=.5129

  
  

• Physics - MathMate, Thursday, November 5, 2009 at 8:38am

  a. use T=2πsqrt(L/g) to get L=g(T/2π)^2 I get .4865 m also. b. At a release angle of 3°, the acceleration is g sin(θ). c. As given in the question, the difference is Δa=(gθ - g sin(θ) ) where θ is measured in radians. I get 0.0002 m/s² d. L(1-sin(θ) I get 0.46. e. Equate energies: mgL(1-sin(&theta))=(1/2)mv² solve for v.

  
  

• Physics-corr - MathMate, Thursday, November 5, 2009 at 5:06pm

  Parts d and e should read: d. L(1-cos(θ) I get 0.000667 m. e. Equate energies: mgL(1-cos(θ))=(1/2)mv² solve for v. v=sqrt(2*9.8*.4865*(1-cos(3°)) I get 0.11 m/s

  
  

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