M3 June 2005 Q6
6. The rise and fall of the water level in a harbour is modelled as simple harmonic motion. On a particular day the maximum and minimum depths of water in the harbour are 10 m and 4 m and these occur at 1100 hours and 1700 hours respectively.
(a) Find the speed, in m h\(^{-1}\), at which the water level in the harbour is falling at 1600 hours on this particular day. (9)
(b) Find the total time, between 1100 hours and 2300 hours on this particular day, for which the depth of water in the harbour is less than 5.5 m. (5)
| Scheme | Marks |
|---|---|
| \(a = 3, \quad T = 12\ \ \left(\text{or } \tfrac{1}{2}T = 6\right)\) | B1, B1 |
| \(T = \dfrac{2\pi}{\omega} = 12 \Rightarrow \omega = \dfrac{\pi}{6}\ \ (\approx 0.52)\) | M1 A1 |
| In the scheme below, when \(a\) and/or \(\omega\) appear in a line, accept the symbols or the candidates’ values of \(a\) and/or \(\omega\) for the marks in that line. | |
| (Taking \(x = a\) when \(t = 0\)) \(x = a\cos\omega t\) | M1 |
| \(\dot{x} = -a\omega\sin\omega t\) | M1 A1 |
| When \(t = 5\) \(\dot{x} = -3 \times \dfrac{\pi}{6}\sin\dfrac{5\pi}{6}\) | M1 |
| \(|\dot{x}| = \dfrac{\pi}{4}\ \ (\text{m h}^{-1})\) awrt 0.79 | A1 |
| (9) |
Notes
The scheme brackets the M1 for \(\dot{x} = -a\omega\sin\omega t\) with the M1 for \(t = 5\): the second depends on the first.
Alternative to 6(a)
| The last 5 marks of 6(a) can be gained as follows. The first 4 marks are as above. | |
| When \(t = 5\) \(x = 3\cos\dfrac{5\pi}{6} = -\dfrac{3\sqrt{3}}{2}\ \ (\approx -2.60)\) | M1 |
| \(v^2 = \omega^2(a^2 - x^2)\) | M1 |
| \(= \dfrac{\pi^2}{6^2}\left(9 - \dfrac{9 \times 3}{4}\right)\ \ \left(= \dfrac{\pi^2}{16}\right)\) | M1 A1 |
| \(|v| = \dfrac{\pi}{4}\ \ (\text{m h}^{-1})\) awrt 0.79 | A1 |
The scheme brackets the M1 for \(v^2 = \omega^2(a^2 - x^2)\) with the next M1: the second depends on the first.
Alternative measuring \(x\) from the centre of oscillation
| (Using 1400 as \(t = 0\)) | |
| The first 4 marks are as above | B1 B1 M1 A1 |
| \(x = a\sin\omega t\) | M1 |
| \(\dot{x} = a\omega\cos\omega t\) | M1 A1 |
| When \(t = 2\) \(\dot{x} = 3 \times \dfrac{\pi}{6}\cos\dfrac{2\pi}{6}\) \(t = 2\) oe is essential for this M | M1 |
| \(= \dfrac{\pi}{4}\ \ (\text{m h}^{-1})\) | A1 |
The scheme brackets the M1 for \(\dot{x}\) with the M1 for \(t = 2\): the second depends on the first.
| Scheme | Marks |
|---|---|
| Depth of 5.5 m \(\Rightarrow x = -1.5\) | |
| \(-1.5 = a\cos\omega t\) | M1 |
| \(\cos\omega t = -\dfrac{1}{2}\) | A1ft |
| \(\dfrac{\pi}{6}t = \dfrac{2\pi}{3},\ \left(\dfrac{4\pi}{3}\right)\) | M1 |
| \(t = 4, 8\) | A1 |
| Required time is \(t_2 - t_1 = 8 - 4 = 4\) (h) | A1 |
| (5) | |
| (14 marks) |
Notes
The scheme brackets the two M1 marks: the second depends on the first.
In 6(b), the following should be accepted
| \(1.5 = a\cos\omega t\) | M1 |
| \(\cos\omega t = \dfrac{1}{2}\) | A1ft |
| \(\dfrac{\pi}{6}t = \dfrac{\pi}{3}\) | M1 |
| \(t = 2\) | A1 |
| Required time is \(2t = 4\) (h) | A1 |
Alternative measuring \(x\) from the centre of oscillation (using 1400 as \(t = 0\))
| \(1.5 = 3\sin\omega t\) | M1 |
| \(\sin\omega t = \dfrac{1}{2}\) | A1ft |
| \(\dfrac{\pi}{6}t = \dfrac{\pi}{6},\ \left(\dfrac{5\pi}{6}\right)\) | M1 |
| \(t = 1, 5\) | A1 |
| Required time is \(t_2 - t_1 = 5 - 1 = 4\) (h) | A1 |
The scheme brackets the two M1 marks: the second depends on the first.