Maximum speed of a string help

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Naldo6
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maximum speed of a string...help please

A tranverse mechanical wave is traveling along a string lying along the x axis. The displacement of the string as function of position and time, y(x,t) is described by the following equiation:

y(x,t)=0.044 sin(4.20x-152t) where x and y are in meters and t is in secnds. Find the wavelength of the wave, the velocity of the wave and the maximum speed in the y direction of any piece of the string.

i have done yet the wavelength and the velocity but i don't know how to calculta the maximum speed in the y direction o any piece of the string.

wavelength= 1.50 meter
velocity of wave= 32.3 m/s
maximum speed= "I don't know"

can anyone help me please
 
on Phys.org


Here are a few pointers:

1) Generally, v= ds/dt, so, what do you need to do with the wavefunction?
2) Where and when does a particle on a wave achieve maximum transverse (in the y direction) velocity ?
 


do yoy mean derivate the y equation?... and set x = o
 


Yes, precisely. There is a simpler method for the 2nd part, however. What is the maximum value of the trig function that you'll find in the derivative?
 


y=0.044 sin(4.20x-152t)

dy/dt= 0.044cos(4.20x-152t)(152)

and then set x=0

is this correct?...
 


not exactly right, what is the maximum value [cos (4.20x-152t)] can take? (Hint: cos(f(x)) has a finite range between ... and ...)
 


i really don't know because in the equation i have a t for time yet, so what i do with that variable?... or what i have to set in the derivate for get the maximum speed?...
 


well, the simple answer is that the maximum value of a cos function is always 1, regardless of the variable present within the function (assuming, of course, that their domains are for all real numbers). For max speed, the cos (4.20x-152t) simply equals 1.
 


oh yes i foget that in an instant so my

y=0.044 sin(4.20x-152t)

dy/dt= 0.044cos(4.20x-152t)(152)

so the max speed is (0.044))(152)

is this now right?... do i derivate correctly?...
 


Yes, that should be right.
 


thanks... it is the correct answer