Answer
(a) $m = 2.22~grams$
(b) $a_{max} = 22400~m/s^2$
Work Step by Step
(a) The expression for the maximum transverse speed at the anti-node is $\omega A$. We can find the angular frequency.
$\omega A = v_{max}$
$\omega = \frac{v_{max}}{A}$
$\omega = \frac{28.0~m/s}{0.0350~m}$
$\omega = 800~rad/s$
We can find the frequency of the first overtone, which is the second harmonic.
$f_2 = \frac{\omega}{2\pi}$
$f_2 = \frac{800~rad/s}{2\pi}$
$f_2 = 127.32~Hz$
We can find the speed of the wave on the string.
$f_2 = \frac{2v}{2L}$
$v = f_2~L$
$v = (127.32~Hz)(2.50~m)$
$v = 318.3~m/s$
We can find the mass of the string.
$v = \sqrt{\frac{F}{\mu}}$
$\mu = \frac{F}{v^2}$
$\frac{m}{L} = \frac{F}{v^2}$
$m = \frac{F~L}{v^2}$
$m = \frac{(90.0~N)(2.50~m)}{(318.3~m/s)^2}$
$m = 0.00222~kg = 2.22~grams$
(b) The expression for the maximum transverse acceleration at the anti-node is $\omega^2A$. We can find the maximum acceleration.
$a_{max} = \omega^2 A$
$a_{max} = (800~rad/s)^2(0.0350~m)$
$a_{max} = 22400~m/s^2$
