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what is your ans?

since we have three complete cycles within 4.8 meters, then the wavelength is:
4.8/3=1.6 meters

since we have 3 waves
so divide 4.8 by 3
because one wave is equal to wavelength

lol
sorry i did not see your post @Michele_Laino

no worries :) @rvc

the requested velocity, is:
4.8/6=...meters/seconds

period is T= 6 /3 =...second
frequency f= 1/T =... Hz

this is how i did it, i first took the inverse of the Ferquency.....F=1/T which agve me the period

how did you get 14.5 meters/second

v=traveled space/time interval= 4.8/6=0.8 m/second

period is
T=6/3= 2 seconds

so the corresponding frequency is:
f=1/T= 1/2=0.5 Hertz

when solving for the very first question wavelength.

i mean second cuz time is given after the wavelength question :(

so do we assume we don't know the time while solving for wavelength

how do i figure out the amplitude? Please help!!!

the picture is at the top.

They are related.

\[\lambda = \frac{ v }{ f }\]

can you quickly guide me what to do Please?!?!?!?!

\[\omega = f 2 \pi\]

It's one in the same, math and physics are interrelated...

Do you have the formula for amplitude?

so use the formula :)

The maximum displacement of the particle from the mean position is the amplitude.

:/

v=n lambda

:)

idk :(

@welshfella please help us

i'll have to look later if you still need it gotta go right now

there are 3 complete waves in the diagram so one wavelength is 4.8 / 3 m

i'll have to check that out.

A = D/F where D is distance travelled by the wave / frequency F

I'm a bit rusty with this stuff.

yes i see hmmm

1.6 m is correct for the wavelength i'm sure of that

D is distance and F = freqency in cycles per second

hint:
in order to find the amplitude, you have to measure the distances D and H as below:

so we have:
\[\Large x = \frac{{4.8 \times H}}{D} = ...\]

what is the distance D? Please measure it with a scale ruler

how many cm is D?
how many cm is H?

ok! and how many cm is D?

what is x?

That's right!
better is x=0.457 meters

it is a proportion

we have finished, since by means of that proportion, our amplitude, is 0.457 meters

yes! I am sure.

yes! since our procedure is correct!

yes! that's right!

thanks!

Thanks @Michele_Laino

http://www.sengpielaudio.com/calculator-speedsound.htm

can you please draw a quick drawing?!?!!?!?? :(

please!?!?!?!?!?!!?!?

|dw:1436132812859:dw|

|dw:1436132987610:dw|

i'll do the next two question after this example, please check them for me, I Love You!!! :")

then we have:
|dw:1436133042674:dw|

i nother words, how do yo know what shape is bigger or smaller :(

the amplitude A is given by this computation:
\[A = \frac{{\max value - \min value}}{2}\]

OMG i'm going to ask you a very sily question but how did you come up with 5?!?!?

I chose those value arbitrarily

values*

oh okay, let me try the next two questions and i'll get back to you.....i hope you'll be here!!! :")

no, you have to measure your numbers using a scale ruler

ok so in what units???? cm?

yes!

ok Thanks you Soooooooooooooooooooooooooooooooooooooooooooooooo MCH!!!! :")

Love you forever!!! :")

Thanks!