The Ferris Wheel Ticket Window
A small fairground runs a Ferris wheel whose passenger cabin height above the ground, in metres, is modelled by h(t) = 12 + 10·sin(πt/15), where t is in seconds after boarding at the platform. The platform is at the lowest point of the ride. Management wants a ticket window placed so that customers can see the cabin rise above a 17-metre safety fence that runs along the boundary. The fence is 40 metres from the wheel's centre line, and the wheel's centre is 12 metres above the ground. A new operator proposes relabelling the model as H(t) = 10·sin(πt/15 − π/2) + 12 to make boarding occur at t = 0 with zero height offset. A surveyor notes that the cabin passes the 17-metre fence height twice per revolution and asks how the timing changes if the platform is raised 2 metres.
1. Determine whether the proposed relabelled function H(t) is equivalent to h(t) for all t, and explain the transformation involved.
Read the worked answer
Compare the two expressions at t = 0.
- h(0) = 12 + 10·sin(0) = 12.
- H(0) = 10·sin(−π/2) + 12 = −10 + 12 = 2.
They differ, so H is not equivalent to h. The phase shift of −π/2 inside the sine moves the graph horizontally, not vertically. Since sin(θ − π/2) = −cos θ, we get H(t) = 12 − 10·cos(πt/15), which starts at the minimum height 2 m, not at 12 m. The original h starts at the mid-height 12 m and rises. So the relabelling changes the physical meaning of t = 0: it would place boarding at the bottom of the ride, contradicting the statement that the platform is at the lowest point only if the platform height is 2 m. If the platform is at 2 m, then H is the correct model and h is wrong; the two models describe different boarding conventions. The judgement call is to check the initial condition, not just the amplitude and period, which are identical (amplitude 10, period 30 s).
2. Find all times in the first revolution when the cabin is exactly at the 17-metre fence height under the original model h(t), and interpret the result physically.
Read the worked answer
Set 12 + 10·sin(πt/15) = 17.
10·sin(πt/15) = 5, so sin(πt/15) = 0.5.
The general solution is πt/15 = π/6 + 2kπ or πt/15 = 5π/6 + 2kπ.
Thus t = 2.5 + 30k or t = 12.5 + 30k seconds.
In the first revolution (0 ≤ t < 30), the cabin reaches 17 m at t = 2.5 s (rising) and t = 12.5 s (falling).
Physical interpretation: the cabin is above the fence height for 10 seconds per revolution, from 2.5 s to 12.5 s. The ticket window must be positioned where sightlines are not blocked during that 10-second window. Note the symmetry: the two times are equidistant from the peak at t = 7.5 s, consistent with the sine graph's axis of symmetry.
3. If the platform is raised by 2 metres, describe how the times found above change, and state whether the answer depends on the amplitude.
Read the worked answer
Raising the platform by 2 m adds 2 to every height: h₂(t) = 14 + 10·sin(πt/15). Solve 14 + 10·sin(πt/15) = 17, so sin(πt/15) = 0.3.
The solutions are πt/15 = arcsin(0.3) ≈ 0.3047 or π − 0.3047 ≈ 2.8369, giving t ≈ 1.455 s and t ≈ 13.545 s.
The cabin now crosses 17 m earlier on the way up and later on the way down, so the above-fence interval widens from 10 s to about 12.09 s. The crossing times depend on the vertical shift relative to the threshold, not on the amplitude alone; changing amplitude would also move the crossings. The key judgement is that a vertical translation changes which fraction of the sine cycle lies above a fixed threshold, so the interval length is not preserved under translation.