# 13.1.1 Shm Simple Pendulums

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Information about 13.1.1 Shm Simple Pendulums
Education

Published on March 24, 2009

Author: christaines

Source: slideshare.net

13.1.1 SHM Simple pendulum

Simple pendulum A pendulum consists of a small “bob” of mass m , suspended by a light inextensible thread of length l , from a fixed point We can ignore the mass of the thread The bob can be made to oscillate about point O in a vertical plane along the arc of a circle

A pendulum consists of a small “bob” of mass m , suspended by a light inextensible thread of length l , from a fixed point

We can ignore the mass of the thread

The bob can be made to oscillate about point O in a vertical plane along the arc of a circle

We can show that oscillating simple pendulums exhibit SHM We need to show that a  x Consider the forces acting on the pendulum: weight , W of the bob and the tension , T in the thread We can resolve W into 2 components parallel and perpendicular to the thread: Parallel: the forces are in equilibrium Perpendicular: only one force acts, providing acceleration back towards O

We can show that oscillating simple pendulums exhibit SHM

We need to show that a  x

Consider the forces acting on the pendulum: weight , W of the bob and the tension , T in the thread

We can resolve W into 2 components parallel and perpendicular to the thread:

Parallel: the forces are in equilibrium

Perpendicular: only one force acts, providing acceleration back towards O

Parallel: F = mg cos  Perpendicular: F = restoring force towards O = mg sin  This is the accelerating force towards O F = ma  - mg sin  = ma (-ve since towards O) When  is small (>10 °) sin    Hence -mg = ma (remember  = s/r = x /l)  -mg ( x /l) = ma Rearranging: a = -g ( x /l) = pendulum equation (can also write this equation as a = - x (g/l) )

Parallel: F = mg cos 

Perpendicular: F = restoring force towards O

= mg sin 

This is the accelerating force towards O

F = ma  - mg sin  = ma (-ve since towards O)

When  is small (>10 °) sin   

Hence -mg = ma (remember  = s/r = x /l)

 -mg ( x /l) = ma

Rearranging: a = -g ( x /l) = pendulum equation

(can also write this equation as a = - x (g/l) )

In SHM a  x Since g/l = constant we can assume a  x for small angles only SHM equation a = -(2  f ) 2 x Pendulum equation a = - x (g/l) Hence (2  f ) 2 = (g/l)  f = 1/2  (  g/l) remember T = 1/ f T = 2  (  l/g) The time period of a simple pendulum depends on length of thread and acceleration due to gravity

In SHM a  x

Since g/l = constant we can assume a  x for small angles only

SHM equation a = -(2  f ) 2 x

Pendulum equation a = - x (g/l)

Hence (2  f ) 2 = (g/l)

 f = 1/2  (  g/l) remember T = 1/ f

T = 2  (  l/g)

The time period of a simple pendulum depends on length of thread and acceleration due to gravity

Measure acceleration of free fall using simple pendulums Use page 36 and 37 of “Physics by Experiment”

Use page 36 and 37 of “Physics by Experiment”

Set up the equipment and set the length of the string so T = 2s Mark a reference point on the stand (to count number of oscillations Displace the pendulum a few centimetres and release – the swing should be 1 plane As the pendulum passes the reference point start the stopwatch and measure the time for 20 oscillations Remember 1 oscillation is from O  A  O  B  O

Set up the equipment and set the length of the string so T = 2s

Mark a reference point on the stand (to count number of oscillations

Displace the pendulum a few centimetres and release – the swing should be 1 plane

As the pendulum passes the reference point start the stopwatch and measure the time for 20 oscillations

Remember 1 oscillation is from O  A  O  B  O

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 Repeat the experiment Record results in a table with the column headings T, T 2 and y (in metres) Plot a graph T 2 against y – this should be a straight line graph

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

Repeat the experiment

Record results in a table with the column headings T, T 2 and y (in metres)

Plot a graph T 2 against y – this should be a straight line graph

Analysis of results T = 2  (  l/g)  T 2 = 4  2 (l/g)  g = 4  2 (l/T 2 ) From your graph find a value for g (from gradient) How does your value of g compare to the accepted value? 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?

T = 2  (  l/g)

 T 2 = 4  2 (l/g)

 g = 4  2 (l/T 2 )

How does your value of g compare to the accepted value?

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?

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