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Simple Harmonic Motion problems

Simple Harmonic Motion problems. A baby is oscillating on a baby bouncer, out of contact with the ground. She has a mass of 5 kg and the bouncer has a spring constant k of 500 Nm -1. What’s the initial extension of the cord?

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Simple Harmonic Motion problems

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  1. Simple Harmonic Motion problems

  2. A baby is oscillating on a baby bouncer, out of contact with the ground. She has a mass of 5 kg and the bouncer has a spring constant k of 500 Nm-1

  3. What’s the initial extension of the cord? • The child is now pulled down a further 0.1 m and released. What is her ANGULAR VELOCITY? • How long after her release does she pass through the equilibrium position? • How fast is she moving at that point? Is it her fastest velocity?

  4. Answers: • Use k = F/ext to show ext = 0.1m •  = (k/m)1/2 = (500/5)1/2 = (100)1/2 = 10 rads-1 • T = 2/10 = 0.63s. T/4 = 0.16 s • V max =  A = 10 x 0.1 = 1ms-1

  5. HIGH and LOW tides.

  6. In a small harbour, the depth of water at high tide is 10 m, and at low tide is zero. The depth of water follows SHM, with amplitude 5 m and periodic time 12 hours (actually 12.5, but round it down to make things simpler). Fishing boats can only leave harbour when the water is 9 m or more. On one day, high tide is at noon. • What is the angular frequency in rad hr-1? • If d is the depth of water ABOVE the mean depth of 5m, write an equation of the form d = A cos t, substituting the right values for  and A. • When is the latest time at which boats can leave the harbour?

  7. Answers: • T = 2/ , so  = 2/T = 1.45 x 10-4 rad s-1 = 0.53 rad h-1 • d = 5 cos (0.53 x t) where t is hours from high tide. • 9 – 5 = 4 = 5 cos (0.53 x t), so t = 1.2 hrs from high tide, which is 1.12 mins after 12 noon, which is 1.12pm. 4 m 5m 9m 5m

  8. Harder question…… Imagine a buoy bobbing up and down in the water…

  9. Displace buoy by x and then let go. Buoy will bob up and down. h L L + x Using the symbols, find an expression for periodic time T of the oscillations of the buoy. You’ll need to use the density of the buoy (d) and of water () & also g.

  10. ANSWER The restoring force on the buoy is  g x A Acceleration = F/m =  g x A / d A h So 2 = a/s =  g x / d h x And  is g / d h Thus T = 2/ = 2 d h/ g cancel m=dxV

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