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
Write this equation in function notation. (1 mark)
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)
T(L) = 2π√(L/9.8)
10 = 2π√(L/9.8)
5/π = √(L/9.8)
25/π^2 = L/9.8
L = 25(9.8)/π^2 = 24.824 m or 24824 mm
change in mass has no effect
To write the equation in function notation, we can use the notation f(L) = 2π√(L/9.8), where f represents the function and L is the input (the length of the pendulum).
The domain of this function is the set of possible input values, which in this case is the length of the pendulum. Since the length cannot be negative (as it represents a physical length), the domain is the set of non-negative real numbers, or [0, ∞).
The range of this function is the set of possible output values, in this case, the time it takes for the pendulum to make one complete swing. As time cannot be negative, the range is also the set of non-negative real numbers, or [0, ∞).
To graph this function, we can plot points using different values of L and their corresponding T values. Let's plot data for pendulums up to 20m in length:
L = 0m, T = f(0) = 2π√(0/9.8) ≈ 0s
L = 1m, T = f(1) = 2π√(1/9.8) ≈ 0.633s
L = 2m, T = f(2) = 2π√(2/9.8) ≈ 0.898s
L = 3m, T = f(3) = 2π√(3/9.8) ≈ 1.146s
L = 4m, T = f(4) = 2π√(4/9.8) ≈ 1.380s
L = 5m, T = f(5) = 2π√(5/9.8) ≈ 1.602s
...and so on.
By plotting these points on a graph with L on the x-axis and T on the y-axis, we can see how the time T changes with different pendulum lengths L up to 20m.
To calculate the length of the pendulum that takes exactly 10s to complete one swing (T = 10s), we can rearrange the equation:
10 = 2π√(L/9.8)
Squaring both sides:
100 = 4π²(L/9.8)
Dividing both sides by 4π²:
L/9.8 = 100 / (4π²) ≈ 2.549
Multiplying both sides by 9.8:
L ≈ 24.950 m
To the nearest millimeter, the length should be approximately 24.950 meters.
If the pendulum were made twice as heavy, its mass would increase but the length of the pendulum would remain the same. The formula for the period of a pendulum does not depend on its mass, only on the length. Therefore, doubling the weight of the pendulum would not affect the value of T.