Answer
(a) f=25 Hz , T=0.04 s , k = 19.6 rad/m
(b) y(x,t) = (0.0700 m)cos[(19.6 rad/m)x+(157 rad/s)t]
(c) y((0.360 m),( 0.150 s)) = 4.94 cm
(d) t=0.005 s
Work Step by Step
(a) Acoording to the formula of the wave speed υ=λf, we can find the frequency f
f=υ/λ=(8.00 m/s)/(0.320 m)=25 Hz
As it's generally known, the time period T is inversely proportional to the frequency f
So T=1/f=1/(25 Hz)=0.04 s
We can find the wave number by using the formula k=2π/λ
k=2π/(0.320 m)= 19.6 rad/m
(b) As the waves travel in the -x direction, the wave equation is: y(x,t)=Acos(kx+ωt+φ)
From the problem's data the amplitude A=0.0700 m
From the (a) k=19.6 rad/m
The angular frequency ω is computed by the formula ω=2πf=2π(25 Hz)=157 rad/s
The φ is the initial phase
The tranverse displacement at x=0 and t=0 is equal to: y(0,0)=Acos(φ)
But, at this point and moment y(0,0) is also equal to A. So Acos(φ)=A<>cos(φ)=1<>cos(φ)=cos(0)<>φ=2κπ
(κ is an integer number). As φ belongs to [0,2π), then κ=0 and φ=0.
So, the equation of the tranverse displacement is: y(x,t)=(0.0700 m)cos[(19.6 rad/m)x+(157 rad/s)t]
(c) At x=0.360 m and t=0.150 s, the tranverse displacement is found by using the formula from (b)
y((0.360 m),( 0.150 s)=(0.0700 m)cos[(19.6 rad/m)(0.360 m)+(157 rad/s)(0.150 s)]=0.0495 m=4.95 cm
(d) At x=0.360 m y(0,t)=(0.0700 m)cos[(157 rad/s)t+7.06 rad] (1)
In order to return the particle at the upper maximum tranverse displacement, we need to equalize the (1) with the 0.0700 m. So cos[(157 rad/s)t+7.06 rad]=1<> cos[(157 rad/s)t+7.06 rad]=cos(0) <> 157t+7.06=2nπ
(n is an integer number and also t>0.150 s). Initially we solve in t, so t=(2nπ-7.06)/157>0.150 (2) <>
2nπ-7.06>157(0.150) <> 2nπ>23.6+7.06 <> 2nπ>30.7 <> n>30.7/2π <> n>4.89
As we need the time for the particle to reach at y=A for the first time after t=0.150s, we will need to take the next possible value for n which is 5. So n=5 and by (2) t=0.155 s
Thus, the requested time t=0.155 s - 0.150 s=0.005 s
