Question

A certain transverse wave is described by y(x,t)=bcos[2π(xl−tτ)], where b = 5.90 mm , l = 28.0 cm , and τ = 3.40×10−2 s .

Answer 1

Part A:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$,

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by $y(x,t)=bcos[2\pi(xl-t\tau)]$, where b = 5.90 mm , l = 28.0 cm , and $\tau = 3.40\times10^{-2} s$.

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:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$,

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by $y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right]$, where b = 5.90 mm , l = 28.0 cm , and $\tau = 3.40\times10^{-2} s$.

Thus,

$y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right[tex] \frac{1}{\lambda} = \frac{1}{l} \\ \\ \Rightarrow\lambda= l =28.0 \ cm=\bold{2.8\times10^{-1}}$

Therefore, the wavelength is 28.0 cm = $2.8\times10^{-1} \ m$.



Part C:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$,

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by $y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right]$, where b = 5.90 mm , l = 28.0 cm , and $\tau = 3.40\times10^{-2} s$.

The wave's frequency, f, is given by:

$f= \frac{1}{T} = \frac{1}{\tau} = \frac{1}{3.40\times10^{-2}} =\bold{29.4 \ Hz}$



Part D:

Given that the the wavelength is $2.8\times10^{-1} \ m$ and that the wave's frequency is 29.4 Hz

The wave's speed of propagation, v, is given by:

$v=f\lambda=29.4(2.8\times10^{-1})=8.232 \ m/s$