At timet = 0and at position x = 0 malong a string, a traveling sinusoidal wave with an angular frequency of 440 rad/shas displacement $\mathit{y}\mathbf{=}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}\mathit{m}$and transverse velocity$\mathit{u}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$ . If the wave has the general form$\mathit{y}\left(x,t\right)\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(kx-\omega t+\varphi \right)$ , what is phase constant $\mathit{\varphi }$?

Angular frequency is$\omega =440rad/s$

Displacement$y=+4.5mm$

Transverse velocity is$u=-0.75m/s$

Wave equation is $y\left(x,t\right)=y_m\sin \left(kx-\omega t+\varphi \right)$

We use the concept of wave motion. We can differentiate the given equation for velocity and take the ratio of displacement by velocity. At initial values of x and t, we get the equation for, plugging the given values, we get the value for phase constant.

Formulae:

The velocity of the body in motion, $v=\frac{dv}{dt}........\left(1\right)$

We know that velocity of the wave using equation (1) is given as:
$$\begin{gathered} u\left(x,t\right)=\frac{dy}{dt} \\ =\frac{d{y}_m\sin \left(kx-\omega t+\varphi \right)}{dt} \\ =-y_m\omega \cos \left(kx-\omega t+\varphi \right) \end{gathered}$$

We know the equation of wave is given as:

$y\left(x,t\right)=y_m\sin \left(kx-\omega t+\varphi \right)$

We can take the ratio ofand

$$\begin{gathered} \frac{y\left(x,t\right)}{u\left(x,t\right)}=\frac{\left(y_m\sin \left(kx-\omega t+\varphi \right)\right)}{-y_m\omega \cos \left(kx-\omega t+\varphi \right)} \\ \frac{y\left(x,t\right)}{u\left(x,t\right)}=\frac{\left(\sin \left(kx-\omega t+\varphi \right)\right)}{-\omega \cos \left(kx-\omega t+\varphi \right)} \\ =\frac{\tan \left(kx-\omega t+\varphi \right)}{\omega } \end{gathered}$$

Plugging the given values in the above equation, we get

$$\begin{gathered} \frac{4.5}{-0.75}=\frac{\tan \left(kx-\omega t+\varphi \right)}{\omega } \\ \mathrm{an}\left(kx-\omega t+\varphi \right)=\frac{4.5\times {10}^{-3}\times 440}{0.75} \\ \left(kx-\omega t+\varphi \right)={\tan }^{-1}\left(2.640\right) \\ \left(kx-\omega t+\varphi \right)=69.25° \end{gathered}$$

For x = 0 then t = 0 , we get the value of phase constant as:

$$\begin{gathered} \varphi =69.25° \\ =69.25°\times \frac{\mathrm{\pi }\mathrm{rad}}{180°} \\ =1.2rad \end{gathered}$$

Hence, the value of phase constant is $1.2rad$.