Question

A wire with mass 50 g is stretched so that it’s ends are tied down at points 100 cm apart. The wire vibrates in its fundamental mode with frequency 60 Hz and an amplitude of 0.5 cm at the antinodes. What is the tension in the wire?

2500 N
720 N
120 N
1250 N

Answer 1

Answer:

720 N

Explanation:

Most of these answers  on here are wrong but I just tool the test ! This is the right answer!

Answer 2

The tension in the wire is 720N

The equation for calculating the wavelength of a standing wave in a string tied at both ends:

                               λ=2L/n

where L is the distance of the tied ends and n is the mode of vibration.

The string is tied at both ends at a distance L= 100cm = 1m. So there will be nodes at both ends. Also the wire is vibrating with fundamental frequency, n =1.

                  λ = 2×1/1 = 2m

now the relation between velocity (v) of the wave and Tension(T) in string is:

                           $v=\sqrt[]{\frac{T}{u} }$

                           ⇒ $T=v^{2}u$

here μ is mass density of string that is 50g/1m = 5×$10^{-2}kg/m$

and velocity,       v = λν  (here ν is the given frequency = 60Hz)

                        ⇒ v = 2×60 = 120 m/s

hence,

                        T = 120×120×5×$10^{-2}$ N

                        T = 720 N

Learn more about standing waves:

https://brainly.com/question/16375282