Question

A standing wave on a string of length L = 3 m fixed at both ends is described by: y(x,t) = 0.01sin⁡(πx)cos⁡(40t), where x and y are in meters and t is in seconds. The maximum transverse speed of an element on the string located at x = 1/6 m is:

Answer 1

Let's first find the transverse speed:

$$v=\dfrac{dy}{dt}=-0.01\ m\times40\ s^{-1}sin(\pi\times\dfrac{1}{6})sin(40t),$$ $$v=-0.2\ \dfrac{m}{s}\times sin(40t).$$

The maximum transverse speed will be when $sin(40t)=\pm1$. Therefore, we get:

$$v_{max}=-0.2\ \dfrac{m}{s}.$$