Skip to content
NCERT Exemplar · Q26

Q.A stationary wave on a stretched string is recorded at two instants of time. At the first instant (t=0t=0) the string is sinusoidal: with the transverse displacement on the vertical axis and position xx (in metres) on the horizontal axis, the string touches the axis (zero displacement) at five equally spaced points labelled, in order along xx, AA, BB, CC, DD and EE. Between consecutive touch points the string bulges alternately upward and downward — a crest in the AA-BB loop, a trough in the BB-CC loop, a crest in the CC-DD loop and a trough in the DD-EE loop. The antinode (point of maximum displacement) in the first loop is labelled A′A' and the antinode in the third loop (between CC and DD) is labelled C′C'. At the second instant the string is completely straight, with zero displacement everywhere. The two waves that superpose to form this stationary wave travel at 360 m/s and each has a frequency of 256 Hz.

(a) Calculate the time at which the second (straight) profile is recorded.
(b) Identify the nodes and the antinodes.
(c) Calculate the distance between A′A' and C′C'.
Sikkim CbseShort· 3mImportance★★★★★est
88% · 51/58 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

With v=360v=360 m/s and f=256f=256 Hz, the wavelength is λ=v/f=1.41\lambda = v/f = 1.41 m and the period is T=1/f=3.9×10−3T = 1/f = 3.9\times10^{-3} s. The string is straight when the time factor cos⁡ωt=0\cos\omega t = 0, first at t=T/4=9.8×10−4t = T/4 = 9.8\times10^{-4} s. Nodes are the fixed axis-crossings A,B,C,D,EA,B,C,D,E; antinodes are the loop-centres A′,C′A', C' (and the other two). A′A' and C′C' are antinodes one full wavelength apart, so A′C′=λ=1.41A'C' = \lambda = 1.41 m.

Given

v=360v = 360 m/s, f=256f = 256 Hz.

λ=vf=360256=1.41 m,T=1f=1256=3.9×10−3 s.\lambda = \frac{v}{f} = \frac{360}{256} = 1.41\ \text{m}, \qquad T = \frac{1}{f} = \frac{1}{256} = 3.9\times10^{-3}\ \text{s}.

(a) Time of the straight profile

A stationary wave is y=2a sin⁡kx cos⁡ωty = 2a\,\sin kx\,\cos\omega t. The string is straight everywhere when the time factor vanishes, cos⁡ωt=0\cos\omega t = 0, first at ωt=π/2\omega t = \pi/2, i.e. t=T/4t = T/4:

t=T4=14×256=9.8×10−4 s.t = \frac{T}{4} = \frac{1}{4\times256} = 9.8\times10^{-4}\ \text{s}.

(b) Nodes and antinodes

Nodes are the permanently stationary points where sin⁡kx=0\sin kx = 0 — the axis-crossings A, B, C, D, EA,\,B,\,C,\,D,\,E. Antinodes are the points of maximum displacement, mid-way between consecutive nodes — the centres of the four loops (the crests/troughs), which include A′A' and C′C'. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.