Question

For the wave shown in figure, if its position shown is at $4l_0{\cos }^2\left(\frac{\pi y}{\beta }\right)$ . Then the equation of the wave is [when speed of wave, $C_1$ ]

• A. $C_2$
• B. $\frac{C_1}{C_2}$
• C. $\frac{a}{b}$
• D. $\frac{2a}{b}$

Answer 1

Answer: (C)

(c.)From the figure $2l_0{\cos }^2\left(\frac{\pi y}{\beta }\right)$ $l_0{\cos }^2\left(\frac{\pi y}{\beta }\right)$ $\frac{l_0}{2}{\cos }^2\left(\frac{2\pi y}{\beta }\right)$ and $4l_0{\cos }^2\left(\frac{\pi y}{\beta }\right)$ At $C_1$ and $C_2$ is $+ve$ Velocity of particle $vp=-v\left(\frac{dy}{dx}\right)=-ve$ Velocity of particle at $x=0$ and $t=0$ downwards. So, initial phase $\varphi =\pi$ So, $y=A\sin (\omega t-kx+\pi )=A\sin (kx-\omega t)$ $=A\sin (5\pi x-2000\pi t)=A\sin 5\pi (x-400t)$