Respuesta :
Answer:
[tex]\lambda_w=0.6509\ m[/tex]
Explanation:
Given:
- mass oscillating with the spring, [tex]m=13.9\ kg[/tex]
- spring constant, [tex]k=9.3\ N.m^{-1}[/tex]
- wave velocity on the water surface, [tex]v_w=5\ m.s^{-1}[/tex]
Now the angular frequency of the spring oscillation:
[tex]\omega=\sqrt{\frac{k}{m} }[/tex]
[tex]\omega=\sqrt{\frac{9.3}{13.9} }[/tex]
[tex]\omega=0.81796\ rad.s^{-1}[/tex]
Now according to the question the wave is created after each cycle of the spring oscillation.
So the time period of oscillation:
[tex]T=\frac{\omega}{2\pi}[/tex]
[tex]T=\frac{0.81796}{2\pi}[/tex]
[tex]T=0.130182\ s[/tex]
Now the wave length of the water wave:
[tex]\lambda_w=v_w.T[/tex]
[tex]\lambda_w=5\times 0.130182[/tex]
[tex]\lambda_w=0.6509\ m[/tex]
Answer:
Wavelength will be 38.388 m
Explanation:
We have given mass m = 13.9 kg
Spring constant K= 9.3 N/m
Velocity v = 5 m /sec
Angular frequency is given by [tex]\omega =\sqrt{\frac{k}{m}}[/tex]
So [tex]\omega =\sqrt{\frac{9.3}{13.9}}=0.817[/tex]
Now we have to find frequency for further calculation
So frequency will be equal to [tex]f=\frac{\omega }{2\pi }=\frac{0.817}{2\times 3.14}=0.130Hz[/tex]
Now we have to find wavelength, it is ratio of velocity and frequency
There is a relation between frequency velocity and wavelength
[tex]v=f\lambda[/tex]
[tex]\lambda =\frac{v}{f}=\frac{5}{0.130}=38.388m[/tex]
