Question

A stretched string has a mass per unit length of 5.12 g/cm and a tension of 19.3 N. A sinusoidal wave on this string has an amplitude of 0.143 mm and a frequency of 76.9 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x,t) = ym sin(kx + ωt), what are (a) ym, (b) k, and (c) ω, and (d) the correct choice of sign in front of ω?

Answer 1

Answer:

a. 0.143 mm b. 77.6 rad/m c. 483.18 rad/s  d. +1

Explanation:

a. ym

Since the amplitude is 0.143 mm, ym = amplitude = 0.143 mm

b. k

We know k = wave number = 2π/λ where λ = wavelength.

Also, λ = v/f where v = speed of wave in string = √(T/μ) where T = tension in string = 19.3 N and μ = mass per unit length = 5.12 g/cm = 5.12 ÷ 1000 kg/(1 ÷ 100 m) = 0.512 kg/m and f = frequency = 76.9 Hz.

So, λ = v/f = √(T/μ)/f

substituting the values of the variables into the equation, we have

λ = √(T/μ)/f

= √(19.6 N/0.512 kg/m)/76.9 Hz

= √(38.28 Nkg/m)/76.9 Hz

= 6.187 m/s ÷ 76.9 Hz

= 0.081 m

= 81 mm

So, k = 2π/λ

= 2π/0.081 m

= 77.6 rad/m

c. ω

ω = angular frequency = 2πf where f = frequency of wave = 76.9 Hz

So, ω = 2πf

= 2π × 76.9 Hz

= 483.18 rad/s

d. The correct choice of sign in front of ω?

Since the wave is travelling in the negative x - direction, the sign in front of ω is positive. That is +1.