Answer to Question #339596 in Physics for sho

Question

A 0.010 0-kg wire, 2.00 m long, is fixed at both ends and vibrates in its

simplest mode under a tension of 200 N. When a vibrating tuning fork is placed

near the wire, a beat frequency of 5.00 Hz is heard. (a) What could be the

frequency of the tuning fork? (b) What should the tension in the wire be if the

beats are to disappear?

Answer 1

(a)

$$f_0=\frac{\sqrt{F/(m/L)}}{2L}=\frac{\sqrt{200/(0.010/2.00)}}{2*2.00}=50\:\rm Hz$$

$$f=f_0\pm\Delta f$$

$$f=50\pm5=55,45\:\rm Hz$$

(b)

$$f=f_0=50\:\rm Hz$$