Respuesta :
Explanation:
(a) It is known that equation for transverse wave is given as follows.
y = [tex](0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)[/tex]
Now, we will compare above equation with the standard form of transeverse wave equation,
y = [tex]A sin(kx + \omega t)[/tex]
where, A is the amplitude = 0.09 m
k is the wave vector = [tex]\frac{\pi}{11}[/tex]
[tex]\omega[/tex] is the angular frequency = [tex]4\pi[/tex]
x is displacement = 1.40 m
t is the time = 0.16 s
Now, we will differentiate the equation with respect to t as follows.
The speed of the wave will be:
v(t) = [tex]\frac{dy}{dt}[/tex]
v(t) = [tex]A \omega cos(kx + \omega t)[/tex]
v(t) = [tex](0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)[/tex]
v(t) = -0.84 m/s
The acceleration of the particle in the location is
a(t) = [tex]\frac{dv}{dt}[/tex]
a(t) = [tex]-A \omega 2sin(kx + \omega t)[/tex]
a(t) = [tex]-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)[/tex]
a(t) = -9.49 [tex]m/s^{2}[/tex]
Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 [tex]m/s^{2}[/tex] .
(b) Wavelength of the wave is given as follows.
[tex]\lambda = \frac{2\pi}{k}[/tex]
[tex]\lambda = (frac{2\pi}{\frac{\pi}{11}) [/tex]
[tex]\lambda[/tex] = 22 m
The period of the wave is
T = [tex]\frac{2 \pi}{\omega}[/tex]
T = [tex]\frac{2 \pi}{4 \pi}[/tex]
= 0.5 sec
Now, we will calculate the speed of propagation of wave as follows.
v = [tex]\frac{\lambda}{T}[/tex]
= [tex]\frac{22 m}{0.5 s}[/tex]
= 44 m/s
therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.
