Respuesta :
Answer:
[tex]T_1=T_3=\dfrac{2\pi}{21}[/tex]
[tex]T_2=T_4=\dfrac{2\pi}{42}[/tex]
Explanation:
Wave 1, [tex]y_1=0.12\ cos(3x-21t)[/tex]
Wave 2, [tex]y_2=0.15\ sin(6x+42t)[/tex]
Wave 3, [tex]y_3=0.13\ cos(6x+21t)[/tex]
Wave 4, [tex]y_4=-0.27\ sin(3x-42t)[/tex]
The general equation of travelling wave is given by :
[tex]y=A\ cos(kx\pm \omega t)[/tex]
The value of [tex]\omega[/tex] will remain the same if we take phase difference into account.
For first wave,
[tex]\omega_1=21[/tex]
[tex]\dfrac{2\pi }{T_1}=21[/tex]
[tex]T_1=\dfrac{2\pi}{21}[/tex]
For second wave,
[tex]\omega_2=42[/tex]
[tex]\dfrac{2\pi }{T_2}=42[/tex]
[tex]T_2=\dfrac{2\pi}{42}[/tex]
For the third wave,
[tex]\omega_3=21[/tex]
[tex]\dfrac{2\pi }{T_3}=21[/tex]
[tex]T_3=\dfrac{2\pi}{21}[/tex]
For the fourth wave,
[tex]\omega_4=42[/tex]
[tex]\dfrac{2\pi }{T_4}=42[/tex]
[tex]T_4=\dfrac{2\pi}{42}[/tex]
It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.
