Posted by James on .
A 48cmlong wire with a mass of 11.0 g is under a tension of 50.5 N. Both ends of the wire are held rigidly while it is plucked.
(a) What is the speed of the waves on the wire?
(b) What is the fundamental frequency of the standing wave?
The answers are 50.5 m/s and 48.9 Hz respectively. How do I get these answers?

Physics 
James,
For part (a) I've been doing the following work:
u=M/L; where u= mass per unit length
u=(0.011)/(.48)= 0.022916667
v= sqrt(T/u); where T = tension
v= sqrt(50.5/0.022916667) = 47.0 m/s 
Physics 
James,
I'm not really sure how to figure out part (b); however, I'm sure that it's answer relies on me figure out part (a) correctly.

Physics 
Elena,
Your answer for part A is correct.
Velocity in the stretched string is
v = sqrt(T/m(o)).
=sqrt(50.5•0.48/0.011)=46.94 m/s
Part B
Part B
λ =2 L= 0.96 m.
λ =v/f,
f=v/ λ =46.94/0.96=48.9 m 
Physics 
James,
Thank you. I guess WebAssign made a mistake.

Physics 
James,
Also, shouldn't f=48.9 m be f=48.9 Hz?

Physics 
Elena,
Certainly, Hz