1504 Speed Of Transverse Wave

Wave Speed on a String

v=Fμ(15.14)\begin{aligned} v = \sqrt{\frac{F}{\mu}} \quad \bold{(15.14)} \end{aligned}

vv - Speed of a transverse wave on a string
FF - Tension in string
μ\mu - Mass per unit length.

The Speed of Mechanical Waves

v=Restoring force returing the system to equilibriumInertia resisting the return to equilibriumv = \sqrt{\frac{\text{Restoring force returing the system to equilibrium}}{\text{Inertia resisting the return to equilibrium}}}

The tension FF in the string plays the role of the restoring force; it tends to bring the string back to its undisturbed, equilibrium configuration. The mass of the string - or, more properly, the linear mass density μ\mu provides the inertia that prevents the string from returning instantaneously to equilibrium. Hence we have v=Fμv =\sqrt{\dfrac{F}{\mu}} for the speed of waves on a string.

Exercises

15, 18, 19

15.15 One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates the rope transversely at 120120 Hz. The other end passes over a pulley and supports a 1.501.50-kg mass. The linear mass density of the rope is 0.0480 kg/m0.0480 \text{ kg/m}.
(a) What is the speed of a transverse wave on the rope?
(b) What is the wavelength?
(c) How would your answers to parts (a) and (b) change if the mass were increased to 3.00 kg?

Solution

v=Fμ=1.5×9.80.048=17.5 m/sλ=vf=17.5120=0.146 m\begin{aligned} v &= \sqrt{\frac{F}{\mu}} = \sqrt{\frac{1.5 \times 9.8}{0.048}} = 17.5 \text{ m/s}\\ \lambda &= \frac{v}{f} = \frac{17.5}{120} = 0.146 \text{ m} \end{aligned}

15.18 A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation

y(x,t)=(8.5 mm)cos⁡(172 rad/mx−4830 rad/st)\begin{aligned} y(x, t) = (8.5 \text{ mm}) \cos(172 \text{ rad/m} x - 4830 \text{ rad/s} t) \end{aligned}

Assume that the tension of the string is constant and equal to WW.
(a) How much time does it take a pulse to travel the full length of the string?
(b) What is the weight WW?
(c) How many wavelengths are on the string at any instant of time?
(d) What is the equation for waves traveling down the string?

Solution

k=2πλ=172 rad/m=172×π180 m−1⇒λ=2πk=2π172×π180=2.09 mω=2πT=4830rad/s=4830×π180 s−1⇒T=2πω=2π4830×π180=0.075 sv=λT=2.090.075=27.9 m/s\begin{aligned} k &=\frac{2\pi}{\lambda} = 172 \text{ rad/m} = 172 \times \frac{\pi}{180} \text{ m}^{-1}\\ \To \lambda &= \frac{2\pi}{k} = \frac{2\pi}{172 \times \frac{\pi}{180}} = 2.09 \text{ m}\\ \omega &= \frac{2\pi}{T} = 4830 \text{rad/s} = 4830 \times \frac{\pi}{180} \text{ s}^{-1}\\ \To T &= \frac{2\pi}{\omega} = \frac{2\pi}{4830 \times \frac{\pi}{180}} = 0.075 \text{ s}\\ v &= \frac{\lambda}{T} = \frac{2.09}{0.075} = 27.9 \text{ m/s}\\ \end{aligned}

15.19 A thin, 75.0-cm wire has a mass of 16.5 g. One end is tied to a nail, and the other end is attached to a screw that can be adjusted to vary the tension in the wire.
(a) To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 cm makes 625 vibrations per second?
(b) How fast would this wave travel?

Solution

μ=16.5 g75.0 cm=0.022 kg/mλ=0.0333 mf=625 Hzv=λf=0.0333×625=20.8 m/sv=Fμ⇒F=v2μ=9.53 N\begin{aligned} \mu &= \frac{16.5 \text { g}}{75.0 \text{ cm}} = 0.022 \text{ kg/m}\\ \lambda &= 0.0333 \text{ m}\\ f &= 625 \text{ Hz}\\ v &= \lambda f = 0.0333 \times 625 = 20.8 \text{ m/s}\\ v &= \sqrt{\frac{F}{\mu}} \To F = v^2 \mu = 9.53 \text{ N} \end{aligned}