Answer
(a) $A = 1.28~cm$
(b) $a_{max} = 6320~m/s^2$
Work Step by Step
(a) We can find the speed of the wave.
$v = \sqrt{\frac{F}{\mu}}$
$v = \sqrt{\frac{F}{(m/L)}}$
$v = \sqrt{\frac{330~N}{(0.00300~kg/2.20~m)}}$
$v = 491.9~m/s$
We can find the frequency.
$f = \frac{v}{2L}$
$f = \frac{491.9~m/s}{(2)(2.20~m)}$
$f = 111.8~Hz$
We can find the angular frequency.
$\omega = 2\pi~f$
$\omega = (2\pi)(111.8~Hz)$
$\omega = 702.5~rad/s$
The expression for the maximum transverse speed at the anti-node is $\omega~A$. We can find the amplitude.
$\omega~A = v_{max}$
$A = \frac{v_{max}}{\omega}$
$A = \frac{9.00~m/s}{(702.5~rad/s)}$
$A = 0.0128~m = 1.28~cm$
(b) The expression for the maximum transverse acceleration at the anti-node is $\omega^2~A$. We can find the maximum acceleration.
$a_{max} = \omega^2~A$
$a_{max} = (702.5~rad/s)^2(0.0128~m)$
$a_{max} = 6320~m/s^2$
