## College Physics (4th Edition)

(a) The maximum transverse speed is $2.86~m/s$ (b) The maximum transverse acceleration of a point on the string is $372~m/s^2$ (c) The wave moves along the string with a speed of $8.67~m/s$ (d) The answer to part (a) is the maximum transverse speed of a point on the string which oscillates perpendicularly to the direction of motion of the wave. The answer to part (c) is the speed of the wave as it moves along the string.
(a) In general: $y(x,t) = A~sin(\omega~t+k~x)$ In this case: $y(x,t) = (2.20~cm)~sin~[~(130~rad/s)~t+(15~rad/m)~x~]$ We can see from the equation that $A = 2.20~cm$ and $\omega = 130~rad/s$. We can find the maximum transverse speed $v_m$: $v_m = A~\omega$ $v_m = (0.0220~m)(130~rad/s)$ $v_m = 2.86~m/s$ The maximum transverse speed is $2.86~m/s$ (b) We can find the maximum transverse acceleration $a_m$ of a point on the string: $a_m = A~\omega^2$ $a_m = (0.0220~m)(130~rad/s)^2$ $a_m = 372~m/s^2$ The maximum transverse acceleration of a point on the string is $372~m/s^2$ (c) We can find the wave speed: $v = \lambda~f$ $v = (\frac{2\pi}{k})(\frac{\omega}{2\pi})$ $v = \frac{\omega}{k}$ $v = \frac{130~rad/s}{15~rad/m}$ $v = 8.67~m/s$ The wave moves along the string with a speed of $8.67~m/s$ (d) The answer to part (a) is the maximum transverse speed of a point on the string which oscillates perpendicularly to the direction of motion of the wave. The answer to part (c) is the speed of the wave as it moves along the string.