Change in water height of a wave pool

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mexqwerty
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A 2.00 m deep swimming pool is equipped with a wave generator that sends sinusoidal waves across the pool. The equation which gives the water depth, h(x,t), some distance x from the wave generator at any time t is:
h(x,t) = 2.00 m + H cos[ 2π [ t/(4.900 s) − x/(0.4000 m) ] − 5π/4 ]
where H = 75.0 cm.

a. What is the change in water height, with respect to the mean water level, a distance 34.81 m from the wave generator at time t = 10.50 s.

b. How much time must elapse from the instant in part (a) until the water 34.81 m from the wave generator reaches its next maximum?


For a, have been trying to do the question and I'm using deltah = H cos[ 2π [ t/(4.900 s) − x/(0.4000 m) ] − 5π/4 ] but obviously its wrong because I'm getting the wrong answer.
 
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mexqwerty said:
A 2.00 m deep swimming pool is equipped with a wave generator that sends sinusoidal waves across the pool. The equation which gives the water depth, h(x,t), some distance x from the wave generator at any time t is:
h(x,t) = 2.00 m + H cos[ 2π [ t/(4.900 s) − x/(0.4000 m) ] − 5π/4 ]
where H = 75.0 cm.

a. What is the change in water height, with respect to the mean water level, a distance 34.81 m from the wave generator at time t = 10.50 s.

b. How much time must elapse from the instant in part (a) until the water 34.81 m from the wave generator reaches its next maximum?


For a, have been trying to do the question and I'm using deltah = H cos[ 2π [ t/(4.900 s) − x/(0.4000 m) ] − 5π/4 ] but obviously its wrong because I'm getting the wrong answer.

You should show details of your actual calculation attempt so that we can see what's going wrong (and it's possible that the "book" answer is incorrect -- it happens sometimes).
 
Oh, never mind. I was doing the right thing but I didn't know you had to set your calculator to radians. Thanks, anyway.
Hmm, but I still don't know how to do the next bit. Do I have to use the equation again? It doesn't look like I can...
 
Last edited:
mexqwerty said:
Oh, never mind. I was doing the right thing but I didn't know you had to set your calculator to radians. Thanks, anyway.

Ah. That'll do it, all right.

Cheers.
 
mexqwerty said:
Hmm, but I still don't know how to do the next bit. Do I have to use the equation again? It doesn't look like I can...
Actually yes, you do use that same equation. What is the value of Δh at a maximum?

p.s. Welcome to Physics Forums.