Answer
(a) $A = 1.72~mm$
(b) $v = 272~m/s$
Work Step by Step
(a) The expression for the maximum transverse speed at the anti-node is $\omega A$. The expression for the maximum transverse acceleration at the anti-node is $\omega^2A$. We can find the angular frequency.
$\frac{\omega^2 A}{\omega A} = \frac{a_{max}}{v_{max}}$
$\omega = \frac{8.40\times 10^3~m/s^2}{3.80~m/s}$
$\omega = 2210.5~rad/s$
We can find the amplitude.
$\omega A = v_{max}$
$A = \frac{v_{max}}{\omega}$
$A = \frac{3.80~m/s}{2210.5~rad/s}$
$A = 0.00172~m = 1.72~mm$
(b) We can find the frequency.
$f_1 = \frac{\omega}{2\pi}$
$f_1 = \frac{2210.5~rad/s}{2\pi}$
$f_1 = 351.8~Hz$
We can find the speed of the transverse waves on the string.
$v = \lambda_1~f_1$
$v = (2L)~f_1$
$v = (2)(0.386~m)(351.8~Hz)$
$v = 272~m/s$
