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# 13.1.1 Shm Simple Pendulums

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### 13.1.1 Shm Simple Pendulums

1. 1. 13.1.1 SHM Simple pendulum
2. 2. Simple pendulum <ul><li>A pendulum consists of a small “bob” of mass m , suspended by a light inextensible thread of length l , from a fixed point </li></ul><ul><li>We can ignore the mass of the thread </li></ul><ul><li>The bob can be made to oscillate about point O in a vertical plane along the arc of a circle </li></ul>
3. 3. <ul><li>We can show that oscillating simple pendulums exhibit SHM </li></ul><ul><li>We need to show that a  x </li></ul><ul><li>Consider the forces acting on the pendulum: weight , W of the bob and the tension , T in the thread </li></ul><ul><li>We can resolve W into 2 components parallel and perpendicular to the thread: </li></ul><ul><li>Parallel: the forces are in equilibrium </li></ul><ul><li>Perpendicular: only one force acts, providing acceleration back towards O </li></ul>
4. 5. <ul><li>Parallel: F = mg cos  </li></ul><ul><li>Perpendicular: F = restoring force towards O </li></ul><ul><li> = mg sin  </li></ul><ul><li>This is the accelerating force towards O </li></ul><ul><li>F = ma  - mg sin  = ma (-ve since towards O) </li></ul><ul><li>When  is small (>10 °) sin    </li></ul><ul><li>Hence -mg = ma (remember  = s/r = x /l) </li></ul><ul><li> -mg ( x /l) = ma </li></ul><ul><li>Rearranging: a = -g ( x /l) = pendulum equation </li></ul><ul><li>(can also write this equation as a = - x (g/l) ) </li></ul>
5. 6. <ul><li>In SHM a  x </li></ul><ul><li>Since g/l = constant we can assume a  x for small angles only </li></ul><ul><li>SHM equation a = -(2  f ) 2 x </li></ul><ul><li>Pendulum equation a = - x (g/l) </li></ul><ul><li>Hence (2  f ) 2 = (g/l) </li></ul><ul><li> f = 1/2  (  g/l) remember T = 1/ f </li></ul><ul><li>T = 2  (  l/g) </li></ul><ul><li>The time period of a simple pendulum depends on length of thread and acceleration due to gravity </li></ul>
6. 7. Measure acceleration of free fall using simple pendulums <ul><li>Use page 36 and 37 of “Physics by Experiment” </li></ul>
7. 8. <ul><li>Set up the equipment and set the length of the string so T = 2s </li></ul><ul><li>Mark a reference point on the stand (to count number of oscillations </li></ul><ul><li>Displace the pendulum a few centimetres and release – the swing should be 1 plane </li></ul><ul><li>As the pendulum passes the reference point start the stopwatch and measure the time for 20 oscillations </li></ul><ul><li>Remember 1 oscillation is from O  A  O  B  O </li></ul>
8. 9. <ul><li>Now change the length of the string (shorter or longer), measuring the length from the point of suspension to the centre of gravity of the bob </li></ul><ul><li>Repeat the experiment </li></ul><ul><li>Record results in a table with the column headings T, T 2 and y (in metres) </li></ul><ul><li>Plot a graph T 2 against y – this should be a straight line graph </li></ul>
9. 10. Analysis of results <ul><li>T = 2  (  l/g) </li></ul><ul><li> T 2 = 4  2 (l/g) </li></ul><ul><li> g = 4  2 (l/T 2 ) </li></ul><ul><li>From your graph find a value for g (from gradient) </li></ul><ul><li>How does your value of g compare to the accepted value? </li></ul><ul><li>This experiment requires that (i) 20 oscillations be timed (ii) the angle of the swing is small (iii) the length of the string is measured to the centre of the bob and (iv) the oscillations are counted as the bob passes the equilibrium point. Why? </li></ul>