Find the value for 'g'.

Planning B

Apparatus:

1. Clamp and stand to support the pendulum.
2. String
3. A suitable mass for the bob of the pendulum.
4. A stopwatch to time the time taken for one oscillation.

Procedure:

1. Set-up the apparatus as shown in the diagram below for a 50cm length.
2. Pull the bob of the pendulum back about 5cm-10cm and let it go.
3. Start the stopwatch and measure the time it takes for twenty oscillations. Then you can divide by twenty to get an accurate average value for the time.
4. Repeat step three to further increase the accuracy of your values.
5. Repeat steps 3 and 4 with lengths of 60cm, 70cm, 80cm, 90cm and 100 cm. And store the values on a table.
6. Use the formula T=2π √(l/g) to then find an approximate value for ‘g’ by plotting a graph. In the graph, plot l vs T2 and draw a line that passes through most of the points.
7. Re arrange the formula above to give g=4π2(l/T2).
8. From the equation, we can see that, (l/T2) is the slope, gradient, of the graph. So we find the gradient of the graph, then we multiply by 4π2 to get the value for ‘g’.
9. Because the length is in cm, we will need to divide our final value for ‘g’ by 100 to get the units of ‘g’ with meters. I.e. instead of having cms-2, we have ms-2, which is more preferred.  

Diagram:

Data Collection

Before starting the experiment; a table, like the one below, can be made to record the data as it is observed.

Table:

Data Presentation

Using the formula T=2π√(l/g) we see that there are no constants being added or subtracted, so when the line has to pass through the origin.

Graph:

As we saw in the graph, the line is of best fit so the value of ‘g’ may vary depending on how I decided to place the line.

The graph passes through the origin because if there was no length, there would be no path for the pendulum to swing and thus we would not be able to calculate a value for ‘g’.

So, taking two points from the graph we can calculate the slope if the graph.

Two points: (0,0) and (1,25)

Gradient of the line = (Δl)-( ΔT2)

So the gradient= (25-0)/(1-0)

So the equation becomes: g=4π2(25)

So ‘g’ is calculated to be approximately 986.96 cms-2.

When we divide by 100 to get ‘g’ in ms-2, we get g ≈ 9.87 ms-2.

Conclusion:

In conclusion, we see that the calculated value of ‘g’ is approximately     9.87 ms-2.

However the textbook value for ‘g’ is 9.80665 ms-2.

The two values are very close and so there may not have been much error.

The procedure was efficient, but could have been made more accurate by involving more lengths e.g. from 10cm to 100cm at intervals of 10cm. Also I could have recorded the time taken for 30 oscillations to obtain further accuracy.

Also instead of drawing a graph, I could have used the equation, g=4π2(l/T2), and substituted the lengths and corresponding times to get different values for ‘g’. Then calculate the average of all the values.