Part A:
The general form of the equation of a transverse wave is given by:
[tex]y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right][/tex]
where A is the amplitude, [tex]\lambda[/tex] is the wavelength, and T is the period.
Given that a certain transverse wave is described by:
[tex]y(x,t)=bcos[2\pi(xl-t\tau)][/tex], where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s
Thus, the amplitude is b = 5.90 mm = 5.9\times10^{-3} \ m
Part B:
The general form of the equation of a transverse wave is given by:
[tex]y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right][/tex]
where A is the amplitude, [tex]\lambda[/tex] is the wavelength, and T is the period.
Given that a certain transverse wave is described by:
[tex]y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right][/tex], where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s
Thus,
[tex]y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right\\ \\ \frac{1}{\lambda} = \frac{1}{l} \\ \\ \Rightarrow\lambda= l =28.0 \ cm=\bold{2.8\times10^{-1}} [/tex]
Part C:
The general form of the equation of a transverse wave is given by:
[tex]y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right][/tex]
where A is the amplitude, [tex]\lambda[/tex] is the wavelength, and T is the period.
Given that a certain transverse wave is described by:
[tex]y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right][/tex],
where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s
The wave's frequency, f, is given by:
[tex]f= \frac{1}{T} = \frac{1}{\tau} = \frac{1}{3.40\times10^{-2}} =\bold{29.4 \ Hz}[/tex]
Part D:
Given that the the wavelength is [tex]2.8\times10^{-1} \ m[/tex] and that the wave's frequency is 29.4 Hz
The wave's speed of propagation, v, is given by:
[tex]v=f\lambda=29.4(2.8\times10^{-1})=8.232 \ m/s[/tex]
