Question

A wire fixed at both ends in 80.0 cm long and has a mass of 1.50 g. The speed of waves on the wire is 320m/s, and it is vibrating in a standing-wave pattern at 800 Hz. (i) What is the distance between adjacent nodes of the standing wave? (ii) Which harmonic does the standing wave present? (iii) What is the tension in the wire? (iv) Determine the frequency when the string vibrates in a standing-wave pattern having eight segments.

Answer 1

Answer:

distance is 0.200 m

harmonic wave is 4th

tension is 192 N

frequency is 64 Hz

Explanation:

given data

mass = 1.50 g

length = 80 cm

speed v = 320 m/s

frequency f = 800 Hz

to find out

distance and Which harmonic wave and  tension and  frequency

solution

we know wavelength here that is

wavelength = v/f = 320 / 800 = 0.400 m

so distance is = wavelength / 2

distance = 0.400 /2 = 0.200 m

and

over string no of half wavelength = length / distance  = 80 / 0.200 =4

so harmonic wave is 4th

and

tension is here

tension = v²×density

tension = 320² × ( 1.50 ×$10^{-3}$ / 80 ×$10^{-2}$ )

tension = 192 N

and

length of string for 8 segment = 8× wavelength /2

wavelength = 80 × 2 / 0.8

wavelength = 0.20 m

frequency = wavelength × v

so frequency = 320 ( 0.20)

frequency = 64 Hz