Question

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 $\mathit{\varphi }$?

Short answer

Phase constant $\varphi$ is 1.2 rad

Step-by-step solution

Given data

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

Displacement$y=+4.5mm$

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

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

Understanding the concept of wave motion

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, $v=\frac{dv}{dt}........\left(1\right)$

Calculation of phase constant

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 $1.2rad$.