Answer
(a) υ=96.0 m/s
(b) T= 461 N
(c) $υ_{y,max}$=1.13 m/s
$a_{y,max}$=4.26 m/s
Work Step by Step
(a) As the ends of the streched wire are tied down at points 80.0 cm apart, it means that the wavelength of the standing wave is the double of its length: λ=2L=2$\times80 cm$=160 cm=1.60 m
We can find the wave speed by using the formula υ=λf=(1.60 m)$\times(60.0 Hz)$=96.0 m/s
(b) The mass of the wire m=40.0 g=40.0$\times$$10^{-3}$ kg= 4$\times$$10^{-2}$ kg
Assuming that the density of the ware μ is stable in all over its length, μ=$\frac{m}{L}$= =$\frac{4\times10^{-2} kg}{0.8 m}$= 0.05 kg/m
In order to find the tension of the wire, we will use the formula that υ=$\sqrt \frac{T}{μ}$
Τhen, T=$υ^{2}$$\times$μ=$(96.0 m/s)^{2}$$\times(0.05 kg/m)$=461 N
(c) The general equation of the displacement of the standing wave is: y(x,t)=Asin(kx)sin(ωt)
Τhe angular frequency is given by the formula: ω=2πf=2π(60.0 Ηz)=377 rad/s
By the problem's data A=0.300 cm=3$\times$$10^{-3}$ m
Here is the equation of the transverse velocity: $υ_{y}$=ωΑsin(kx)cos(ωt)
So, the maximum transverse velocity is:$υ_{y}$=ωΑ=(377 rad/s)(3$\times$$10^{-3}$ m)= 1.13 m/s
Also, the equation of the transverse acceleration is: $a_{y}$=-$ω^{2}$Αsin(kx)sin(ωt)
Thus, the maximum transverse acceleration is: $a_{y}$=$ω^{2}$Α=$(377 rad/s)^{2}$$\times$3$\times$$10^{-3}$ m =
