The length of time that it takes for a pendulum to make one complete swing depends on the length of the pendulum. The time in seconds, T, is related to the length in metres, L, through the equation T=2π√ L/9.8
State the domain and range of this function. (1 mark)
Graph this function for pendulums up to 20m in length. (3 marks)
If you wanted to construct a pendulum in the Ontario Science Centre that takes exactly 10s to complete one swing, how long would it have to be (to the nearest millimetre)? (2 marks)
What would be the effect on the value of T if the pendulum were made twice as heavy? (1 mark)
Can you please answer all the question.
Please can you guys help with the graph.
all positive L and ll positive T
I am not going to try to draw a graph here. Put in a few values for L say L = 0, 1, 5, 20, 30 and calculate T
10 = 2 pi sqrt (L/g)
1.5915 = sqrt(L/9.8)
L/9.8 = 2.533
L = 24.824
To the nearest millimeter makes no sense by he way because 9.8 is a very rough approximation for g
To find the domain and range of the function T = 2π√(L/9.8), we need to consider the restrictions on the variables L and T.
Domain:
The domain represents the possible values for the independent variable L (length of the pendulum). In this case, the length cannot be negative. Additionally, the function is only defined for real numbers, so the length should not be a complex number. The domain of the function is therefore L ≥ 0.
Range:
The range represents the possible values for the dependent variable T (time in seconds). Since T represents the time, it cannot be negative. Additionally, the square root of a non-negative number will always yield a non-negative result. Therefore, the range of the function is T ≥ 0.
Now let's move on to graphing the function for pendulums up to 20m in length.
To create a graph, we can plot the length (L) on the x-axis and the time (T) on the y-axis. We will only consider lengths up to 20m, as mentioned in the question.
Here's a step-by-step process to create the graph:
1. Choose a suitable scale for the x-axis and y-axis. For example, you can use increments of 1 meter for the x-axis and 1 second for the y-axis.
2. Select various lengths (L) within the range of 0 to 20 meters. For example, you can choose L = 2m, 5m, 10m, 15m, and 20m.
3. Plug these chosen lengths into the equation T = 2π√(L/9.8) to calculate the corresponding times (T).
For example, when L = 2m:
T = 2π√(2/9.8) = 2.004 seconds (approximately)
Repeat this calculation for the other chosen lengths.
4. Plot the corresponding points (L, T) on the graph.
5. Connect the points with a smooth curve to show the relationship between L and T.
6. Extend the graph up to L = 20m.
Now let's answer the question about constructing a pendulum that takes exactly 10 seconds to complete one swing at the Ontario Science Centre.
To find the length of the pendulum (L) that gives a time (T) of 10 seconds, we can rearrange the equation: T = 2π√(L/9.8) and solve for L.
1. Start with the given equation: 10 = 2π√(L/9.8)
2. Square both sides of the equation to remove the square root: 100 = 4π²(L/9.8)
3. Simplify the equation: 100 = (4π²/9.8)L
4. Solve for L: L = (100 * 9.8) / (4π²) ≈ 7.848 meters
Therefore, the pendulum at the Ontario Science Centre would need to be approximately 7.848 meters long.
Regarding the effect on the value of T if the pendulum were made twice as heavy:
Since the equation T = 2π√(L/9.8) only involves the length of the pendulum (L), doubling the weight or mass of the pendulum does not affect the time (T). The relationship between T and L remains the same regardless of the weight.