# Wave function - displacement - transverse wave

## Homework Statement

A transverse traveling wave on a string starts at x = 0 and travels towards x = ∞. The wave has an amplitude of 1.20 m, wavelength of 4.60 m and travels at a speed of 14.3 m/s . At time t = 0.0 s the displacement at position x = 0.0 m is 1.20 m.
(b) Calculate the displacement at position x = 2.6 m when t = 8.60 s.

An identical wave is launched on a new string, whose mass per unit length doubles at some position x > 0.
(c) Determine wavefunctions for the reflected and transmitted waves. You do not need to consider their amplitudes.

## Homework Equations

y(x,t)=Acos(kx-wt)

## The Attempt at a Solution

I think since at x=0 and t=0 , the displacement is equal to 1.2m, then we need to use y(x,t)=Acos(kx-wt) not y(x,t)=Asin(kx-wt) please tell me if this is true.
I got an answer for the displacement y(x,t)=y(2.6,8.60)=0,495m, using this formula: y(x,t)=Acos(kx-wt) but I am not sure if this is true. Please tell me if I am wrong and what should I do to fix it.
I found out that f=3.11Hz w=19.54 rad/s k=1.36

Orodruin
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y(x,t)=Acos(kx-wt) not y(x,t)=Asin(kx-wt) please tell me if this is true.
The general expression is ##y(x,t) = A \sin(kx - \omega t + \phi)##, where ##\phi## is a constant. You can fix this constant using your information. The equations you quote are special cases. Does your choice satisfy the special case?

I found out that f=3.11Hz w=19.54 rad/s k=1.36
How did you arrive at these conclusions? It helps us if you write out your argumentation and computations so that they can be checked. What relations did you use to arrive at these numbers?

The general expression is ##y(x,t) = A \sin(kx - \omega t + \phi)##, where ##\phi## is a constant. You can fix this constant using your information. The equations you quote are special cases. Does your choice satisfy the special case?

How did you arrive at these conclusions? It helps us if you write out your argumentation and computations so that they can be checked. What relations did you use to arrive at these numbers?
f=v/lamda=14.3/4.6=3.11 Hz
k=2pi/lamda=2pi/4.6=1.36

this is given:
A=1.20 m
t= 8.60 s
x = 2.6

y(x,t)=Acos(kx-wt)
y(x,t)=1.2cos(1.36x2.6-19.64x8.6)
y(x,t)=1.2(-0.42)
y(x,t)=-0.51m

i think i was wrong with my previous result, but I am still unsure with the one that I got now

haruspex