(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v\), frequency \(f\), amplitude \(A\), and wavelength \(\lambda \). Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) \(x = \frac{\lambda }{2}\), (ii) \(x = \frac{\lambda }{4}\), and (iii) \(x = \frac{\lambda }{8}\), from the left-hand end of the string.

(b) At each of the points in part (a), what is the amplitude of the motion?

(c) At each of the points in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement?

The given data can be listed below as,

• The speed is, \(v\).
• The frequency is, \(f\).
• The amplitude is, \(A\).
• The wavelength is, \(\lambda \).

This velocity is perpendicular to the wave's direction of propagation for a transverse wave. By calculating the velocity's partial derivative with respect to time, which is the position's second time derivative, we were able to determine the acceleration.

\({a_y} = \frac{{{\partial ^2}y}}{{\partial {t^2}}}\) …(1)

The standard equation of the standing wave is expressed as,

\(y\left( {x,t} \right) = \left( {{A_{sw}}\sin kx} \right)\sin \omega t\) …(2)

Here \({A_{sw}}\) is the amplitude of simple harmonic, \(k\) is the wave number and \(\omega \) is the angular frequency.

But,

The relation between wave number and wavelength is expressed as,

\(k = \frac{{2\pi }}{\lambda }\) …(3)

Here \(\lambda \) is the wavelength.

Substitute the value of \(k\) in equation ()

\(y\left( {x,t} \right) = \left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda }x} \right)\sin \omega t\) …(4)

For a velocity,

\({v_y} = \frac{{\partial y}}{{\partial t}}\)

Equation (3) differentiate with respect to \(t\).

\({v_y} = \frac{{\partial y}}{{\partial t}}\)

\({v_y} = \omega \left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda }x} \right)\cos \omega t\) …(5)

For maximum velocity,

\({\left( {{v_y}} \right)_{\max }} = \omega \left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda }x} \right)\) …(6)

\(\begin{array}{l}{\left( {{v_y}} \right)_{\max }} = \omega \left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda } \times \frac{\lambda }{2}} \right)\\{\left( {{v_y}} \right)_{\max }} = 0\end{array}\)

At \(x = \frac{\lambda }{2}\) is a node and there is no motion.

At \(x = \frac{\lambda }{4}\)

Substitute \(x = \frac{\lambda }{4}\) in equation (6)

\(\begin{array}{c}{\left( {{v_y}} \right)_{\max }} = \omega \left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda } \times \frac{\lambda }{4}} \right)\\{\left( {{v_y}} \right)_{\max }} = A\omega \\ = A\left( {2\pi f} \right)\\{\left( {{v_y}} \right)_{\max }} = 2\pi fA\end{array}\)

At \(x = \frac{\lambda }{4}\) is an antinode and maximum velocity is, \({\left( {{v_y}} \right)_{\max }} = 2\pi fA\).

Substitute \(x = \frac{\lambda }{8}\) in equation (6)

\(\begin{array}{c}{\left( {{v_y}} \right)_{\max }} = \omega \left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda } \times \frac{\lambda }{8}} \right)\\{\left( {{v_y}} \right)_{\max }} = A\omega \left( {\frac{1}{{\sqrt 2 }}} \right)\\ = A\left( {2\pi f} \right)\left( {\frac{1}{{\sqrt 2 }}} \right)\\{\left( {{v_y}} \right)_{\max }} = \sqrt 2 \pi fA\end{array}\)

At \(x = \frac{\lambda }{8}\), the maximum velocity is, \({\left( {{v_y}} \right)_{\max }} = \sqrt 2 \pi fA\).

For an acceleration,

\({a_y} = \frac{{{\partial ^2}y}}{{\partial {t^2}}}\)

Equation (5) differentiate with respect to \(t\).

\({a_y} = - {\omega ^2}\left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda }x} \right)\sin \omega t\) …(7)

For maximum acceleration,

\({\left( {{a_y}} \right)_{\max }} = {\omega ^2}\left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda }x} \right)\) …(8)

Substitute \(x = \frac{\lambda }{2}\) in equation (8)

\(\begin{array}{l}{\left( {{a_y}} \right)_{\max }} = {\omega ^2}\left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda } \times \frac{\lambda }{2}} \right)\\{\left( {{a_y}} \right)_{\max }} = 0\end{array}\)

At \(x = \frac{\lambda }{2}\) the maximum acceleration is zero.

Substitute \(x = \frac{\lambda }{4}\) in equation (8)

\(\begin{array}{l}{\left( {{a_y}} \right)_{\max }} = {\omega ^2}\left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda } \times \frac{\lambda }{4}} \right)\\{\left( {{a_y}} \right)_{\max }} = {\omega ^2}A\\{\left( {{a_y}} \right)_{\max }} = {\left( {2\pi f} \right)^2}A\\{\left( {{a_y}} \right)_{\max }} = 4{\pi ^2}{f^2}A\end{array}\)

At \(x = \frac{\lambda }{4}\) the maximum acceleration is, \({\left( {{a_y}} \right)_{\max }} = 4{\pi ^2}{f^2}A\).

Substitute \(x = \frac{\lambda }{8}\) in equation (8)

\(\begin{array}{l}{\left( {{a_y}} \right)_{\max }} = {\omega ^2}\left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda } \times \frac{\lambda }{8}} \right)\\{\left( {{a_y}} \right)_{\max }} = {\omega ^2}A \times \left( {\frac{1}{{\sqrt 2 }}} \right)\\{\left( {{a_y}} \right)_{\max }} = {\left( {2\pi f} \right)^2}A \times \left( {\frac{1}{{\sqrt 2 }}} \right)\\{\left( {{a_y}} \right)_{\max }} = 2\sqrt 2 {\pi ^2}{f^2}A\end{array}\)

At \(x = \frac{\lambda }{8}\) the maximum acceleration is, \({\left( {{a_y}} \right)_{\max }} = 2\sqrt 2 {\pi ^2}{f^2}A\).

Hence,

1. At \(x = \frac{\lambda }{2}\) is a node and there is no motion, the maximum velocity and maximum acceleration are zero
2. At \(x = \frac{\lambda }{4}\) is an antinode and maximum velocity is, \({\left( {{v_y}} \right)_{\max }} = 2\pi fA\) and the maximum acceleration is, \({\left( {{a_y}} \right)_{\max }} = 4{\pi ^2}{f^2}A\).
3. At \(x = \frac{\lambda }{8}\) the maximum velocity is, \({\left( {{v_y}} \right)_{\max }} = \sqrt 2 \pi fA\) and the maximum acceleration is, \({\left( {{a_y}} \right)_{\max }} = 2\sqrt 2 {\pi ^2}{f^2}A\).