Answer:
Amplitude, A = 0.0205 m
Frequency, f = 0.636 Hz
Wavelength, 18.84 m
Explanation:
Given that,
Speed of the wave along a taught string, v = 12 m/s
A fixed point on the string oscillates as a function of time according to the equation :
[tex]y = 0.0205\ \cos(4t)[/tex]........(1)
Here, y is the displacement in meters and the time t is in seconds
The general equation of oscillation is given by :
[tex]y=A\ \cos\omega t[/tex]............(2)
On comparing equation (1) and (2) we get :
(a) Amplitude, A = 0.0205 m
(b) Frequency of the wave,
[tex]\omega=4\\\\2\pi f=4\\\\f=\dfrac{4}{2\pi}\\\\f=0.636\ Hz[/tex]
(c) Velocity of the wave is given by formula as :
[tex]v=\dfrac{\omega}{k}[/tex]
Since,
[tex]k=\dfrac{2\pi}{\lambda}[/tex]
[tex]v=\dfrac{\omega\lambda}{2\pi}\\\\\lambda=\dfrac{2\pi v}{\omega}\\\\\lambda=\dfrac{2\pi \times 12}{4}\\\\\lambda=18.84\ m[/tex]
Hence, this is the required solution.
