Determining metres that a frequency produces

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Homework Statement


The end of a 256 Hz tuning fork has an average amplitude of 1.0 mm for 2.0 seconds. How many metres does the end of the tuning fork travel in this time.

(The answer is also given in the back of the textbook, which is 2.0m).


Homework Equations


v = f * lambda, I believe.
f = cycles/seconds


The Attempt at a Solution


Well this is what I'm thinking. One wavelength would be 4mm. To the left, rest position, right, and back at resposition. (Each time the fork vibrates left & right it goes 1mm out).

Therefore 1 cycle = 4mm (0.004m)
(2/0.004m)
So, would the period be 500 metres per s?

So how many metres does the end of the tuning fork travel in 2 seconds?
Well.. wouldn't it be 1000 metres?

The answer says 2metres though. I think I don't have an understanding of the question. If someone could explain, I would very much appreciate it. Thank you so much!
 

Discussion

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Your number of 4 mm per cycle is correct. For this tuning fork, how many cycles are there in 2 seconds?
 
No frequency is cycles per second. How do you get 0.25?
 
Another name for 1 Hz is "one cycle per second."