# Shm

#### Air

Simple Harmonic Motion

Question:
On a particular day, high tide at the entrance to a harbour occurs at 11a.m. and the water depth is 14m. Low tide occurs 6.25 hours later at which time the water depth is 6m. In a model of the situation, the water level is assumed to perform simple harmonic motion.Using this model,

(a) write down the amplitude and period of the motion.

A ship needs a depth of 9m before it can enter or leave the harbour.
(b) Show that on this day a ship must enter the harbour by 2.38p.m., correct to the nearest minute, or wait for low tide to pass.

My Solution:
(a) Amplitude = 0.5(14 − 6) = 4m, Period = 2 × 6.25 = 12.5 hours

My Problem:
I can't do part (b). Can someone help with it please. Thanks in advance.

Last edited:

#### topsquark

Forum Staff
Question:
On a particular day, high tide at the entrance to a harbour occurs at 11a.m. and the water depth is 14m. Low tide occurs 6.25 hours later at which time the water depth is 6m. In a model of the situation, the water level is assumed to perform simple harmonic motion.Using this model,

(a) write down the amplitude and period of the motion.

A ship needs a depth of 9m before it can enter or leave the harbour.
(b) Show that on this day a ship must enter the harbour by 2.38p.m., correct to the nearest minute, or wait for low tide to pass.

My Solution:
(a) Amplitude = 0.5(14 − 6) = 4m, Period = 2 × 6.25 = 12.5 hours

My Problem:
I can't do part (b). Can someone help with it please. Thanks in advance.
What is the equation for the water level then? The average height of the water is
$$\displaystyle \frac{14~m + 6~m}{2} = 10~m$$
so
$$\displaystyle d = 10 + 4~cos \left ( \frac{2 \pi}{12.5} t \right )$$
(taking our 0 for time at 11:00 AM, t is measured in hours, and d is measured in m.)

So the boat can only get in while the water is above 9 m. When does the water level drop to 9 m then? Solve
$$\displaystyle 9 = 10 + 4~cos \left ( \frac{2 \pi}{12.5} t \right )$$
for t.

-Dan

Air

#### Air

How did you know that it was $$\displaystyle x=A\cos (\omega t)$$ and not $$\displaystyle x=A\sin (\omega t)$$?

#### topsquark

Forum Staff
How did you know that it was $$\displaystyle x=A\cos (\omega t)$$ and not $$\displaystyle x=A\sin (\omega t)$$?
The wave peaked at 11:00 AM. Cosine peaks at 0. So I simply defined my time variable to start at 11:00 AM and used cosine.

We could just as easily used
$$\displaystyle sin \left ( \frac{2 \pi}{12.5} t + \frac{\pi}{2} \right )$$
which is equal to the cosine

or we could have been trickier and started our clocks at noon. That would mean that 11:00 AM would be at t = -1 hours and make (the full) equation
$$\displaystyle d = 10 + 4~cos \left ( \frac{2 \pi}{12.5} (t + 1) \right )$$

-Dan

Air