Question

doubt related to wave function

Answer 1

a wave pulse is travelling in the +x direction on a string at 2m/s. Displacement y (in cm) of the particle at x=0 at any time t is given by \(\frac{2}{(t^2 +1)}\). Find the expression of the function y=(x,t), t.t., displacement of a particle at position x and time t.

Answer 2

replacing t by t- x/v where v is 2 solves the problem but what is the logic behind this replacement?

well if x/v is the time period of the wave then it is fine but x/v is just the time taken to travel a distance x with a velocity v and not necessarily the time period right?

Answer 3

please note that, when I write:
\[f\left( {x,t} \right) = f\left( {x,t - \frac{x}{v}} \right)\]
I'm saying that the value of the function \(f\) at time \(t\), is equal to its value at time \(t-x/v\), which is a preceding time with respect to \(t\)

Answer 4

okay
but how is the value of function same at t and t-x/v

Answer 5

since, a wave function, is a \(periodic\) function, please think about the function which describes th electric field of a plane electromagnetic wave

Answer 6

we can choose \(x\), in \(t-x/v\), proportionally to the wave length

Answer 7

why didn't we replace t with t+x/v ?

Answer 8

yes a wave function is periodic
and the value of a periodic function repeats after a regular interval called time period so does that mean that x/v is the time period?

Answer 9

second answer is correct!
first answer:
because if I write:
\[f\left( {x,t} \right) = f\left( {x,t + \frac{x}{v}} \right)\]
I'm saying that the value of \(f\) at time \(t\), is equal to its value at a subsequent time \(t+x/v\), so I'm describing a backward wave

Answer 10

i have done a similar problem before in which this concept was used-

y(x,t)=y(x-vt,0)

i understand this one
here after t=0 the wave is moving towards right side
and after time=t it will cover vt distance so the new x coordinate will be (x-vt)+vt=x

so basically y(x,t) and y(x-vt,0) are representing the same y coordinate on the wave

Answer 11

yes! Correct!

Answer 12

so x/v is the time period?

Answer 13

yes! we can choose \(x=v\;T\), where \(T\) is the period of the wave

Answer 14

time period is the time in which a wave travels a distance equal to its wave length
so does that mean that x is the wavelength :/

Answer 15

yes! It is not a contradiction

Answer 16

we can see better the reasoning, if we write this:
\[f\left( x \right) = f\left( {x - vt} \right)\quad \left( {{\text{progressive}}\;{\text{wave}}} \right)\]

Answer 17

more precisely:
\[f\left( {x,t} \right) = f\left( {x - vt,t} \right)\quad \left( {{\text{progressive}}\;{\text{wave}}} \right)\]

Answer 18

but the question says -
" Find the expression of the function y=(x,t), t.t., displacement of a particle at position x and time t"

so x here indicates the displacement of particle
so in our expression x must indicate the displacement of particle

our expression after making that replacement is this-

\(\frac{2}{(t-x/2)^2+1}\) and here we said that x is the wavelength and x/2 the time period

but according to the question x is the displacement and it can take any value
and not every x/2 is the wavelength so x/2 cannot be the time period

Answer 19

please think about this reasoning:
let's consider a progressive wave:
\(f(x-vt)\)
and the subsequent drawing:

wherein a wave at two different times, \(t_0,t\) is depicted. Now the value of \(f\), at positions \(x',x''\) is equal, if:
\[\begin{gathered}
x' - v{t_0} = x'' - vt \hfill \\
\hfill \\
\frac{{x'' - x'}}{{t - {t_0}}} = v \hfill \\
\end{gathered} \]
namely the Whole wave \(f\) is moving towards the right side