Types of Waves — Question 4

Section 16-1 · Physics pages 2398–2399

Question 4 A single pulse need not possess one wavelength A rope carries one isolated triangular pulse of width w = .40 m moving at v = 2.0 m/s without changing shape. There is no repeating train of pulses. x v = 2.0 m/s w = 0.40 m Pulse geometry and direction; the profile is schematic unless a function is given. (a) Find the duration for which a fixed point is disturbed. (b) Explain why labeling w as the wavelength and v/w as an intrinsic frequency is unjustified.
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Question 4 - Solution A single pulse need not possess one wavelength x v = 2.0 m/s w = 0.40 m Method and mathematical working The detector begins responding when the leading edge reaches it and stops when the trailing edge passes. The relevant distance is the pulse width, not the distance between periodic repetitions. ttrailing −tleading = xleading −xtrailing v = w v . ∆t = 0.40 m 2.0 m/s = 0.20 s, ∆t̸ = T for an isolated pulse. (a) Calculation and conclusion The leading and trailing edges are separated by w, so the duration at a fixed point is w/v = .20 s. This is a passage duration, not a repetition period. (b) Interpretation and limits A wavelength is the spatial repeat distance of a periodic pattern; this isolated pulse has none. Likewise, no repeating temporal cycle defines a unique frequency. The scale v/w = 5 s−1 is an inverse duration, and a spectral description would involve multiple frequencies. Physical check Doubling the pulse width at fixed speed doubles the passage duration. This limiting comparison works without assigning a period or wavelength to an isolated pulse.

Question 4

A single pulse need not possess one wavelength

A rope carries one isolated triangular pulse of width w=.40mw=.40\,\mathrm{m} moving at v=2.0m/sv=2.0\,\mathrm{m/s} without changing shape. There is no repeating train of pulses.

Diagram shown in the original SVG worksheet.

  1. Find the duration for which a fixed point is disturbed.

  2. Explain why labeling ww as the wavelength and v/wv/w as an intrinsic frequency is unjustified.

Question 4 - Solution

A single pulse need not possess one wavelength

Diagram shown in the original SVG worksheet.

Method and mathematical working

The detector begins responding when the leading edge reaches it and stops when the trailing edge passes. The relevant distance is the pulse width, not the distance between periodic repetitions.ttrailing−tleading=xleading−xtrailingv=wv.Δt=0.40m2.0m/s=0.20s,Δt≠Tfor an isolated pulse.\begin{aligned}t_{\mathrm{trailing}}-t_{\mathrm{leading}}&=\frac{x_{\mathrm{leading}}-x_{\mathrm{trailing}}}{v}=\frac wv.\\[4pt]\Delta t&=\frac{0.40\,\mathrm m}{2.0\,\mathrm{m/s}}=0.20\,\mathrm s,\qquad \Delta t\ne T\ \text{for an isolated pulse}.\end{aligned}

(a) Calculation and conclusion

The leading and trailing edges are separated by ww, so the duration at a fixed point is w/v=.20sw/v=.20\,\mathrm{s}. This is a passage duration, not a repetition period.

(b) Interpretation and limits

A wavelength is the spatial repeat distance of a periodic pattern; this isolated pulse has none. Likewise, no repeating temporal cycle defines a unique frequency. The scale v/w=5s−1v/w=5\,\mathrm{s^{-1}} is an inverse duration, and a spectral description would involve multiple frequencies.

Physical check

Doubling the pulse width at fixed speed doubles the passage duration. This limiting comparison works without assigning a period or wavelength to an isolated pulse.