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

4. Answers: a)Use k = F/ext to show ext = 0.1m b)  = (k/m) 1/2 = (500/5) 1/2 = (100) 1/2 = 10 rads -1 c)T = 2  /10= 0.63s. T/4 = 0.16 s d)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. a)What is the angular frequency in rad hr -1 ? b)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. c)When is the latest time at which boats can leave the harbour?

7. Answers: a)T = 2  / , so  = 2  /T = 1.45 x 10 -4 rad s -1 = 0.53 rad h -1 b)d = 5 cos (0.53 x t) where t is hours from high tide. c) 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. 5m 9m 4 m

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

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

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