A stretched string has a mass per unit length of 5.0 g/cm and a tension of 15 N. A wave on this string has an amplitude of 0.13 mm and a frequency of 200 Hz and is traveling in the positive x direction. (a) What is the wave speed? (b) Write an equation for this wave (c) What is the wavelength of the traveling wave? (d) What is the maximum speed of a particle on the string?
The Correct Answer and Explanation is:
We are given:
- Mass per unit length: μ = 5.0 g/cm = 0.005 kg/m
- Tension: T = 15 N
- Frequency: f = 200 Hz
- Amplitude: A = 0.13 mm = 1.3 × 10⁻⁴ m
(a) What is the wave speed?
The wave speed vvv on a string is given by:v=Tμv = \sqrt{\frac{T}{\mu}}v=μT
Substitute the values:v=150.005=3000≈54.77 m/sv = \sqrt{\frac{15}{0.005}} = \sqrt{3000} ≈ 54.77\ \text{m/s}v=0.00515=3000≈54.77 m/s
(b) Write an equation for this wave
A traveling wave moving in the positive x-direction can be written as:y(x,t)=Asin(kx−ωt)y(x, t) = A \sin(kx – \omega t)y(x,t)=Asin(kx−ωt)
Where:
- A=1.3×10−4A = 1.3 \times 10^{-4}A=1.3×10−4 m
- f=200f = 200f=200 Hz
- ω=2πf=400π rad/s\omega = 2\pi f = 400\pi \ \text{rad/s}ω=2πf=400π rad/s
- v=54.77 m/sv = 54.77\ \text{m/s}v=54.77 m/s
- λ=vf=54.77200≈0.274 m\lambda = \frac{v}{f} = \frac{54.77}{200} ≈ 0.274\ \text{m}λ=fv=20054.77≈0.274 m
- k=2πλ=2π0.274≈22.94 rad/mk = \frac{2\pi}{\lambda} = \frac{2\pi}{0.274} ≈ 22.94\ \text{rad/m}k=λ2π=0.2742π≈22.94 rad/m
So, the wave equation is:y(x,t)=(1.3×10−4)sin(22.94x−400πt)y(x, t) = (1.3 \times 10^{-4}) \sin(22.94x – 400\pi t)y(x,t)=(1.3×10−4)sin(22.94x−400πt)
(c) What is the wavelength?
λ=vf=54.77200≈0.274 m\lambda = \frac{v}{f} = \frac{54.77}{200} ≈ 0.274\ \text{m}λ=fv=20054.77≈0.274 m
(d) What is the maximum speed of a particle on the string?
The maximum particle speed is:vmax=ωA=(400π)(1.3×10−4)≈0.163 m/sv_{\text{max}} = \omega A = (400\pi)(1.3 \times 10^{-4}) ≈ 0.163\ \text{m/s}vmax=ωA=(400π)(1.3×10−4)≈0.163 m/s
Explanation
This problem explores the motion of a transverse wave on a stretched string. First, we calculate the wave speed using the formula v=T/μv = \sqrt{T/\mu}v=T/μ. The tension in the string provides the restoring force for wave motion, while the mass per unit length resists acceleration. A higher tension results in faster waves, while a higher mass per unit length slows them down. With a tension of 15 newtons and a linear mass density of 0.005 kilograms per meter, the wave speed is found to be approximately 54.77 meters per second.
Next, we describe the wave mathematically. A sinusoidal traveling wave in the positive x-direction is represented as y(x,t)=Asin(kx−ωt)y(x, t) = A \sin(kx – \omega t)y(x,t)=Asin(kx−ωt). The amplitude AAA is how far particles on the string move up and down from equilibrium, here converted to meters. The angular frequency ω\omegaω is related to the frequency by ω=2πf\omega = 2\pi fω=2πf, and the wave number kkk is calculated from the wavelength as k=2π/λk = 2\pi/\lambdak=2π/λ.
To find the wavelength, we divide the wave speed by the frequency. This gives the distance between two adjacent crests or troughs. Lastly, the maximum speed of a particle on the string is given by vmax=ωAv_{\text{max}} = \omega Avmax=ωA, which indicates how fast a point on the string is moving up and down at its peak. This value is different from the wave speed, which is the speed of energy transfer along the string.
