91影视

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.

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

The standard equation of the standing wave is expressed as,

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

\(k = \frac{{2\pi }}{\lambda }\) 鈥�(3)

Here \(\lambda \) is the wavelength.

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

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

For a velocity,

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

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

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

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

For maximum velocity,

\({\left( {{v_y}} \right)_{\max }} = \omega \left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda }x} \right)\) 鈥�(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 \(t\).

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

For maximum acceleration,

\({\left( {{a_y}} \right)_{\max }} = {\omega ^2}\left( {{A_{sw}}\sin \frac{{2\pi }}{\lambda }x} \right)\) 鈥�(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, \({\left( {{a_y}} \right)_{\max }} = 2\sqrt 2 {\pi ^2}{f^2}A\).