Question

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents the displacement. I: y = 0.12 cos(3x - 21t) II: y = 0.15 sin(6x + 42t) III: y = 0.13 cos(6x + 21t) IV: y = -0.27 sin(3x - 42t) Which of these waves have the same period?

Answer 1

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

$T_2=T_4=\dfrac{2\pi}{42}$

Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

Wave 3, $y_3=0.13\ cos(6x+21t)$

Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

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.