Physics · Ch 11 — Waves

Velocity of Transverse Waves in a Stretched String

11.3.1

Velocity of Transverse Waves in a Stretched String

The speed of a transverse wave travelling along a stretched string can be derived by applying Newton's second law to a tiny curved element of the string as a wave pulse passes through it. Consider two nearby points A and B on the string, separated by a short elemental length $dl$ of mass $dm=\mu\,dl$ , where $\mu=dm/dl$ is the string's linear mass density (mass per unit length). As the pulse passes, this small element momentarily traces the arc of a circle of radius $R$ (centred at some point O), subtending a small angle $\theta=dl/R$ there; viewed from a reference frame travelling along with the pulse at speed $v$ , the element needs a centripetal force $F_{cp}=(dm)v^2/R$ directed toward the arc's centre to keep it moving along this momentary curved path. That centripetal force is supplied by the string's own tension $T$ , which acts tangentially at points A and B: resolving the tension at each end into horizontal and vertical components shows the horizontal components cancel exactly by symmetry, while the vertical components (each approximately $T\sin(\theta/2)\approx T\theta/2$ for small $\theta$ ) add together to give a net inward radial force $F_r = T\theta = T\,dl/R$ . Equating this net restoring force to the required centripetal force, $T\,dl/R = \mu(dl)v^2/R$ , the arc length $dl$ and the radius $R$ both cancel completely from both sides, leaving the clean …

Figure 11.15Elemental segment in a stretched string, zoomed, with the pulse seen from an observer frame moving with velocity v

What this figure shows. A short curved element of a string, between two nearby points A and B, is shown zoomed in and magnified, drawn as a small arc of a circle of radius R centred at a point O, subtending a small angle theta at O. Tension forces T are drawn acting tangentially at A and at B, each resolved into a horizontal component $T\cos(\theta/2)$ and a vertical component $T\sin(\theta/2)$ pointing toward the centre O; a label V (or v) marks the pulse's own travelling speed relative to a stationary observer, while the whole diagram is drawn from the point of view of a reference frame moving along with the pulse itself, in which frame the element simply appears to sweep along the arc under this net inward (centripetal) force. This figure is the entire geometric basis for the derivation that follows: the horizontal tension components at A and B cancel by symmetry, but the vertical components add up to supply exactly the centripetal force needed to keep the small mass element $dm=\mu\,dl$ moving along its momentary circular arc, and equating that net force to $dm,v …

Misc Example 11.6Speed and travel time of a pulse on a hanging-mass string

Worked out. A string carries a travelling pulse and has a linear mass density of 0.25 kg/m, and a mass of 1.2 kg is hung from it to provide the tension; the task is to find the pulse's speed and the time it takes to cover 30 cm along the string. The tension supplied by the hanging mass is $T = mg = 1.2\times9.8 = 11.76\ \text{N}$ , so applying the string wave-speed formula gives $v=\sqrt{T/\mu}=\sqrt{11.76/0.25}=\sqrt{47.04}\approx6.86\ \text{m/s}$ . The time to cover a distance of 30 cm (0.30 m) then follows from $t=d/v=0.30/6.86\approx0.044\ \text{s}=44\ \text{ms}$ , showing how directly a measured hanging weight and a known string density let the pulse speed, and hence its travel time over any given length, b …

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