Chapter 11, Problem 101P

To determine

The graphs of the transverse velocity and acceleration of the point $x=0$ as functions of time showing one complete cycle.

Answer to Problem 101P

The graph of the transverse velocity of the point $x=0$ as function of time is

And that of acceleration is

Explanation of Solution

The given wave is harmonic.

Write the general expression for a harmonic wave.

$y\left(x,t\right)=A\sin \left(\omega t+kx\right)$ (I)

Here, $y\left(x,t\right)$ is the instantaneous displacement of the particles of the medium, $A$ is the amplitude, $\omega$ is the angular velocity, $t$ is the time, $k$ is the wave vector and $x$ is the distance travelled by the wave

Write the expression of the given standing wave.

$y\left(x,t\right)=\left(2.00\text{ mm}\right)\sin \left[\left(157\text{ rad/s}\right)t+\left(7.85\text{ rad/m}\right)x\right]$ (II)

Compare equations (I) and (II).

$\begin{array}{c}A=2.00\text{ mm}\\ \omega =157\text{ rad/s}\\ k=7.85\text{ rad/m}\end{array}$

Write the expression for the maximum velocity.

$v_m=\omega A$

Here, $v_m$ is the maximum velocity

Substitute $157\text{ rad/s}$ for $\omega$ and $2.00\text{ mm}$ for $A$ in the above equation to find $v_m$ .

$\begin{array}{c}{v}_m=\left(157\text{ rad/s}\right)\left(2.00\text{ mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)\right)\\ =\left(157\text{ rad/s}\right)\left(0.002\text{ m}\right)\\ =0.314\text{ m/s}\end{array}$

Transverse velocity leads $y\left(x,t\right)$ by a quarter cycle, so that it must be a cosine function.

Write the equation for transverse velocity.

$v_y=v_m\cos \left(\omega t+kx\right)$

Here, $v_y$ is the transverse velocity

Substitute $0.314\text{ m/s}$ for $v_m$ , $157\text{ rad/s}$ for $\omega$ and $2.00\text{ mm}$ for $A$ in the above equation to find $v_y$ .

$v_y=\left(0.314\text{ m/s}\right)\cos \left[\left(157\text{ rad/s}\right)t+\left(7.85\text{ rad/m}\right)x\right]$ (III)

Write the equation for the maximum acceleration for a point on the string.

$a_m={\omega }^{2}A$

Here, $a_m$ is the maximum acceleration

Substitute $157\text{ rad/s}$ for $\omega$ and $2.00\text{ mm}$ for $A$ in the above equation to find $a_m$ .

$\begin{array}{c}{a}_m={\left(157\text{ rad/s}\right)}^2\left(2.00\text{ mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)\right)\\ ={\left(157\text{ rad/s}\right)}^2\left(0.002\text{ m}\right)\\ =49.3{\text{ m/s}}^2\end{array}$

Transverse acceleration leads $v_y$ by a quarter cycle so that it leads $y\left(x,t\right)$ by a half cycle. This implies transverse acceleration is a negative sine function.

Write the equation for the transverse acceleration.

$a_y=-a_m\sin \left(\omega t+kx\right)$

Here, $a_y$ is the transverse acceleration

Substitute $49.3{\text{ m/s}}^2$ for $a_m$ , $157\text{ rad/s}$ for $\omega$ and $2.00\text{ mm}$ for $A$ in the above equation to find $a_y$ .

$a_y=-\left(49.3{\text{ m/s}}^2\right)\sin \left[\left(157\text{ rad/s}\right)t+\left(7.85\text{ rad/m}\right)x\right]$ (IV)

Write the equation for the time period of the wave.

$T=\frac{2\pi }{\omega }$

Here, $T$ is the time period of the wave

Substitute $157\text{ rad/s}$ for $\omega$ in the above equation to find $T$ .

$\begin{array}{c}T=\frac{2\pi }{157\text{ rad/s}}\\ =40.0\times {10}^{-3}\text{ s}\\ =40.0\text{ ms}\end{array}$

Conclusion:

Substitute $0$ for $x$ in equation (III).

$\begin{array}{c}{v}_y=\left(0.314\text{ m/s}\right)\cos \left[\left(157\text{ rad/s}\right)t+\left(7.85\text{ rad/m}\right)0\right]\\ =\left(0.314\text{ m/s}\right)\cos \left[\left(157\text{ rad/s}\right)t\right]\end{array}$

The graph of the above equation is plotted in figure 1.

Substitute $0$ for $x$ in equation (IV).

$\begin{array}{c}{a}_y=-\left(49.3{\text{ m/s}}^2\right)\sin \left[\left(157\text{ rad/s}\right)t+\left(7.85\text{ rad/m}\right)0\right]\\ =-\left(49.3{\text{ m/s}}^2\right)\sin \left[\left(157\text{ rad/s}\right)t\right]\end{array}$

The graph of the above equation is plotted in figure 2.

Thus, the graphs of the transverse velocity and acceleration of the point $x=0$ as functions of time showing one complete cycle are plotted.